Calculus Examples

$f (x) = 12 x^{5} - 75 x^{4} + 160 x^{3} - 120 x^{2} + 4$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $12 x^{5} - 75 x^{4} + 160 x^{3} - 120 x^{2} + 4$ with respect to $x$ is $\frac{d}{d x} [12 x^{5}] + \frac{d}{d x} [- 75 x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [12 x^{5}] + \frac{d}{d x} [- 75 x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 1.2

Evaluate $\frac{d}{d x} [12 x^{5}]$ .

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Step 1.2.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x^{5}$ with respect to $x$ is $12 \frac{d}{d x} [x^{5}]$ .

$12 \frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 75 x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$12 (5 x^{4}) + \frac{d}{d x} [- 75 x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 1.2.3

Multiply $5$ by $12$ .

$60 x^{4} + \frac{d}{d x} [- 75 x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 1.3

Evaluate $\frac{d}{d x} [- 75 x^{4}]$ .

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Step 1.3.1

Since $- 75$ is constant with respect to $x$ , the derivative of $- 75 x^{4}$ with respect to $x$ is $- 75 \frac{d}{d x} [x^{4}]$ .

$60 x^{4} - 75 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$60 x^{4} - 75 (4 x^{3}) + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 1.3.3

Multiply $4$ by $- 75$ .

$60 x^{4} - 300 x^{3} + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 1.4

Evaluate $\frac{d}{d x} [160 x^{3}]$ .

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Step 1.4.1

Since $160$ is constant with respect to $x$ , the derivative of $160 x^{3}$ with respect to $x$ is $160 \frac{d}{d x} [x^{3}]$ .

$60 x^{4} - 300 x^{3} + 160 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 1.4.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$60 x^{4} - 300 x^{3} + 160 (3 x^{2}) + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 1.4.3

Multiply $3$ by $160$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 1.5

Evaluate $\frac{d}{d x} [- 120 x^{2}]$ .

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Step 1.5.1

Since $- 120$ is constant with respect to $x$ , the derivative of $- 120 x^{2}$ with respect to $x$ is $- 120 \frac{d}{d x} [x^{2}]$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 120 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$

Step 1.5.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 120 (2 x) + \frac{d}{d x} [4]$

Step 1.5.3

Multiply $2$ by $- 120$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x + \frac{d}{d x} [4]$

Step 1.6

Differentiate using the Constant Rule.

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Step 1.6.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x + 0$

Step 1.6.2

Add $60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$ and $0$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$ with respect to $x$ is $\frac{d}{d x} [60 x^{4}] + \frac{d}{d x} [- 300 x^{3}] + \frac{d}{d x} [480 x^{2}] + \frac{d}{d x} [- 240 x]$ .

$f'' (x) = \frac{d}{d x} (60 x^{4}) + \frac{d}{d x} (- 300 x^{3}) + \frac{d}{d x} (480 x^{2}) + \frac{d}{d x} (- 240 x)$

Step 2.2

Evaluate $\frac{d}{d x} [60 x^{4}]$ .

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Step 2.2.1

Since $60$ is constant with respect to $x$ , the derivative of $60 x^{4}$ with respect to $x$ is $60 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = 60 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 300 x^{3}) + \frac{d}{d x} (480 x^{2}) + \frac{d}{d x} (- 240 x)$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f'' (x) = 60 (4 x^{3}) + \frac{d}{d x} (- 300 x^{3}) + \frac{d}{d x} (480 x^{2}) + \frac{d}{d x} (- 240 x)$

Step 2.2.3

Multiply $4$ by $60$ .

$f'' (x) = 240 x^{3} + \frac{d}{d x} (- 300 x^{3}) + \frac{d}{d x} (480 x^{2}) + \frac{d}{d x} (- 240 x)$

Step 2.3

Evaluate $\frac{d}{d x} [- 300 x^{3}]$ .

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Step 2.3.1

Since $- 300$ is constant with respect to $x$ , the derivative of $- 300 x^{3}$ with respect to $x$ is $- 300 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = 240 x^{3} - 300 \frac{d}{d x} x^{3} + \frac{d}{d x} (480 x^{2}) + \frac{d}{d x} (- 240 x)$

Step 2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = 240 x^{3} - 300 (3 x^{2}) + \frac{d}{d x} (480 x^{2}) + \frac{d}{d x} (- 240 x)$

Step 2.3.3

Multiply $3$ by $- 300$ .

$f'' (x) = 240 x^{3} - 900 x^{2} + \frac{d}{d x} (480 x^{2}) + \frac{d}{d x} (- 240 x)$

Step 2.4

Evaluate $\frac{d}{d x} [480 x^{2}]$ .

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Step 2.4.1

Since $480$ is constant with respect to $x$ , the derivative of $480 x^{2}$ with respect to $x$ is $480 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 240 x^{3} - 900 x^{2} + 480 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 240 x)$

Step 2.4.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 240 x^{3} - 900 x^{2} + 480 (2 x) + \frac{d}{d x} (- 240 x)$

Step 2.4.3

Multiply $2$ by $480$ .

$f'' (x) = 240 x^{3} - 900 x^{2} + 960 x + \frac{d}{d x} (- 240 x)$

Step 2.5

Evaluate $\frac{d}{d x} [- 240 x]$ .

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Step 2.5.1

Since $- 240$ is constant with respect to $x$ , the derivative of $- 240 x$ with respect to $x$ is $- 240 \frac{d}{d x} [x]$ .

$f'' (x) = 240 x^{3} - 900 x^{2} + 960 x - 240 \frac{d}{d x} x$

Step 2.5.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 240 x^{3} - 900 x^{2} + 960 x - 240 \cdot 1$

Step 2.5.3

Multiply $- 240$ by $1$ .

$f'' (x) = 240 x^{3} - 900 x^{2} + 960 x - 240$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

$\frac{d}{d x} [12 x^{5}] + \frac{d}{d x} [- 75 x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 4.1.2

Evaluate $\frac{d}{d x} [12 x^{5}]$ .

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Step 4.1.2.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x^{5}$ with respect to $x$ is $12 \frac{d}{d x} [x^{5}]$ .

$12 \frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 75 x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 4.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$12 (5 x^{4}) + \frac{d}{d x} [- 75 x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 4.1.2.3

Multiply $5$ by $12$ .

$60 x^{4} + \frac{d}{d x} [- 75 x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 4.1.3

Evaluate $\frac{d}{d x} [- 75 x^{4}]$ .

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Step 4.1.3.1

Since $- 75$ is constant with respect to $x$ , the derivative of $- 75 x^{4}$ with respect to $x$ is $- 75 \frac{d}{d x} [x^{4}]$ .

$60 x^{4} - 75 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 4.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$60 x^{4} - 75 (4 x^{3}) + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 4.1.3.3

Multiply $4$ by $- 75$ .

$60 x^{4} - 300 x^{3} + \frac{d}{d x} [160 x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 4.1.4

Evaluate $\frac{d}{d x} [160 x^{3}]$ .

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Step 4.1.4.1

Since $160$ is constant with respect to $x$ , the derivative of $160 x^{3}$ with respect to $x$ is $160 \frac{d}{d x} [x^{3}]$ .

$60 x^{4} - 300 x^{3} + 160 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 4.1.4.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$60 x^{4} - 300 x^{3} + 160 (3 x^{2}) + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 4.1.4.3

Multiply $3$ by $160$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} + \frac{d}{d x} [- 120 x^{2}] + \frac{d}{d x} [4]$

Step 4.1.5

Evaluate $\frac{d}{d x} [- 120 x^{2}]$ .

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Step 4.1.5.1

Since $- 120$ is constant with respect to $x$ , the derivative of $- 120 x^{2}$ with respect to $x$ is $- 120 \frac{d}{d x} [x^{2}]$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 120 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$

Step 4.1.5.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 120 (2 x) + \frac{d}{d x} [4]$

Step 4.1.5.3

Multiply $2$ by $- 120$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x + \frac{d}{d x} [4]$

Step 4.1.6

Differentiate using the Constant Rule.

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Step 4.1.6.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x + 0$

Step 4.1.6.2

Add $60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$ and $0$ .

$f' (x) = 60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$

Step 5

Set the first derivative equal to $0$ then solve the equation $60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x = 0$

Step 5.2

Factor the left side of the equation.

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Step 5.2.1

Factor $60 x$ out of $60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$ .

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Step 5.2.1.1

Factor $60 x$ out of $60 x^{4}$ .

$60 x (x^{3}) - 300 x^{3} + 480 x^{2} - 240 x = 0$

Step 5.2.1.2

Factor $60 x$ out of $- 300 x^{3}$ .

$60 x (x^{3}) + 60 x (- 5 x^{2}) + 480 x^{2} - 240 x = 0$

Step 5.2.1.3

Factor $60 x$ out of $480 x^{2}$ .

$60 x (x^{3}) + 60 x (- 5 x^{2}) + 60 x (8 x) - 240 x = 0$

Step 5.2.1.4

Factor $60 x$ out of $- 240 x$ .

$60 x (x^{3}) + 60 x (- 5 x^{2}) + 60 x (8 x) + 60 x (- 4) = 0$

Step 5.2.1.5

Factor $60 x$ out of $60 x (x^{3}) + 60 x (- 5 x^{2})$ .

$60 x (x^{3} - 5 x^{2}) + 60 x (8 x) + 60 x (- 4) = 0$

Step 5.2.1.6

Factor $60 x$ out of $60 x (x^{3} - 5 x^{2}) + 60 x (8 x)$ .

$60 x (x^{3} - 5 x^{2} + 8 x) + 60 x (- 4) = 0$

Step 5.2.1.7

Factor $60 x$ out of $60 x (x^{3} - 5 x^{2} + 8 x) + 60 x (- 4)$ .

$60 x (x^{3} - 5 x^{2} + 8 x - 4) = 0$

Step 5.2.2

Factor $x^{3} - 5 x^{2} + 8 x - 4$ using the rational roots test.

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Step 5.2.2.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 4, \pm 2$

$q = \pm 1$

Step 5.2.2.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 4, \pm 2$

Step 5.2.2.3

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

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Step 5.2.2.3.1

Substitute $1$ into the polynomial.

$1^{3} - 5 \cdot 1^{2} + 8 \cdot 1 - 4$

Step 5.2.2.3.2

Raise $1$ to the power of $3$ .

$1 - 5 \cdot 1^{2} + 8 \cdot 1 - 4$

Step 5.2.2.3.3

Raise $1$ to the power of $2$ .

$1 - 5 \cdot 1 + 8 \cdot 1 - 4$

Step 5.2.2.3.4

Multiply $- 5$ by $1$ .

$1 - 5 + 8 \cdot 1 - 4$

Step 5.2.2.3.5

Subtract $5$ from $1$ .

$- 4 + 8 \cdot 1 - 4$

Step 5.2.2.3.6

Multiply $8$ by $1$ .

$- 4 + 8 - 4$

Step 5.2.2.3.7

Add $- 4$ and $8$ .

$4 - 4$

Step 5.2.2.3.8

Subtract $4$ from $4$ .

$0$

Step 5.2.2.4

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 5 x^{2} + 8 x - 4}{x - 1}$

Step 5.2.2.5

Divide $x^{3} - 5 x^{2} + 8 x - 4$ by $x - 1$ .

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Step 5.2.2.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{3}$

$5 x^{2}$

$8 x$

$4$

Step 5.2.2.5.2

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$

Step 5.2.2.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			+	$x^{3}$	-	$x^{2}$

Step 5.2.2.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - x^{2}$

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$

Step 5.2.2.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$

Step 5.2.2.5.6

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$

Step 5.2.2.5.7

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$4 x$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$

Step 5.2.2.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					-	$4 x^{2}$	+	$4 x$

Step 5.2.2.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x^{2} + 4 x$

				$x^{2}$	-	$4 x$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	-	$4 x$

Step 5.2.2.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$4 x$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	-	$4 x$
							+	$4 x$

Step 5.2.2.5.11

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	-	$4 x$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	-	$4 x$
							+	$4 x$	-	$4$

Step 5.2.2.5.12

Divide the highest order term in the dividend $4 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	-	$4 x$
							+	$4 x$	-	$4$

Step 5.2.2.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	-	$4 x$
							+	$4 x$	-	$4$
							+	$4 x$	-	$4$

Step 5.2.2.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $4 x - 4$

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	-	$4 x$
							+	$4 x$	-	$4$
							-	$4 x$	+	$4$

Step 5.2.2.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$8 x$	-	$4$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	-	$4 x$
							+	$4 x$	-	$4$
							-	$4 x$	+	$4$
										$0$

Step 5.2.2.5.16

Since the remander is $0$ , the final answer is the quotient.

$x^{2} - 4 x + 4$

Step 5.2.2.6

Write $x^{3} - 5 x^{2} + 8 x - 4$ as a set of factors.

$60 x ((x - 1) (x^{2} - 4 x + 4)) = 0$

Step 5.2.3

Factor.

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Step 5.2.3.1

Factor using the perfect square rule.

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Step 5.2.3.1.1

Rewrite $4$ as $2^{2}$ .

$60 x ((x - 1) (x^{2} - 4 x + 2^{2})) = 0$

Step 5.2.3.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Step 5.2.3.1.3

Rewrite the polynomial.

$60 x ((x - 1) (x^{2} - 2 \cdot x \cdot 2 + 2^{2})) = 0$

Step 5.2.3.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

$60 x ((x - 1) {(x - 2)}^{2}) = 0$

Step 5.2.3.2

Remove unnecessary parentheses.

$60 x (x - 1) {(x - 2)}^{2} = 0$

Step 5.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 1 = 0$

${(x - 2)}^{2} = 0$

Step 5.4

Set $x$ equal to $0$ .

$x = 0$

Step 5.5

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 5.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 5.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5.6

Set ${(x - 2)}^{2}$ equal to $0$ and solve for $x$ .

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Step 5.6.1

Set ${(x - 2)}^{2}$ equal to $0$ .

${(x - 2)}^{2} = 0$

Step 5.6.2

Solve ${(x - 2)}^{2} = 0$ for $x$ .

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Step 5.6.2.1

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 5.6.2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 5.7

The final solution is all the values that make $60 x (x - 1) {(x - 2)}^{2} = 0$ true.

$x = 0, 1, 2$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = 0, 1, 2$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$240 {(0)}^{3} - 900 {(0)}^{2} + 960 (0) - 240$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

Raising $0$ to any positive power yields $0$ .

$240 \cdot 0 - 900 {(0)}^{2} + 960 (0) - 240$

Step 9.1.2

Multiply $240$ by $0$ .

$0 - 900 {(0)}^{2} + 960 (0) - 240$

Step 9.1.3

Raising $0$ to any positive power yields $0$ .

$0 - 900 \cdot 0 + 960 (0) - 240$

Step 9.1.4

Multiply $- 900$ by $0$ .

$0 + 0 + 960 (0) - 240$

Step 9.1.5

Multiply $960$ by $0$ .

$0 + 0 + 0 - 240$

Step 9.2

Simplify by adding and subtracting.

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Step 9.2.1

Add $0$ and $0$ .

$0 + 0 - 240$

Step 9.2.2

Add $0$ and $0$ .

$0 - 240$

Step 9.2.3

Subtract $240$ from $0$ .

$- 240$

Step 10

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 11

Find the y-value when $x = 0$ .

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Step 11.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = 12 {(0)}^{5} - 75 {(0)}^{4} + 160 {(0)}^{3} - 120 {(0)}^{2} + 4$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Raising $0$ to any positive power yields $0$ .

$f (0) = 12 \cdot 0 - 75 {(0)}^{4} + 160 {(0)}^{3} - 120 {(0)}^{2} + 4$

Step 11.2.1.2

Multiply $12$ by $0$ .

$f (0) = 0 - 75 {(0)}^{4} + 160 {(0)}^{3} - 120 {(0)}^{2} + 4$

Step 11.2.1.3

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 - 75 \cdot 0 + 160 {(0)}^{3} - 120 {(0)}^{2} + 4$

Step 11.2.1.4

Multiply $- 75$ by $0$ .

$f (0) = 0 + 0 + 160 {(0)}^{3} - 120 {(0)}^{2} + 4$

Step 11.2.1.5

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 + 0 + 160 \cdot 0 - 120 {(0)}^{2} + 4$

Step 11.2.1.6

Multiply $160$ by $0$ .

$f (0) = 0 + 0 + 0 - 120 {(0)}^{2} + 4$

Step 11.2.1.7

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 + 0 + 0 - 120 \cdot 0 + 4$

Step 11.2.1.8

Multiply $- 120$ by $0$ .

$f (0) = 0 + 0 + 0 + 0 + 4$

Step 11.2.2

Simplify by adding numbers.

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Step 11.2.2.1

Add $0$ and $0$ .

$f (0) = 0 + 0 + 0 + 4$

Step 11.2.2.2

Add $0$ and $0$ .

$f (0) = 0 + 0 + 4$

Step 11.2.2.3

Add $0$ and $0$ .

$f (0) = 0 + 4$

Step 11.2.2.4

Add $0$ and $4$ .

$f (0) = 4$

Step 11.2.3

The final answer is $4$ .

$y = 4$

Step 12

Evaluate the second derivative at $x = 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$240 {(1)}^{3} - 900 {(1)}^{2} + 960 (1) - 240$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

One to any power is one.

$240 \cdot 1 - 900 {(1)}^{2} + 960 (1) - 240$

Step 13.1.2

Multiply $240$ by $1$ .

$240 - 900 {(1)}^{2} + 960 (1) - 240$

Step 13.1.3

One to any power is one.

$240 - 900 \cdot 1 + 960 (1) - 240$

Step 13.1.4

Multiply $- 900$ by $1$ .

$240 - 900 + 960 (1) - 240$

Step 13.1.5

Multiply $960$ by $1$ .

$240 - 900 + 960 - 240$

Step 13.2

Simplify by adding and subtracting.

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Step 13.2.1

Subtract $900$ from $240$ .

$- 660 + 960 - 240$

Step 13.2.2

Add $- 660$ and $960$ .

$300 - 240$

Step 13.2.3

Subtract $240$ from $300$ .

$60$

Step 14

$x = 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 1$ is a local minimum

Step 15

Find the y-value when $x = 1$ .

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Step 15.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = 12 {(1)}^{5} - 75 {(1)}^{4} + 160 {(1)}^{3} - 120 {(1)}^{2} + 4$

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

One to any power is one.

$f (1) = 12 \cdot 1 - 75 {(1)}^{4} + 160 {(1)}^{3} - 120 {(1)}^{2} + 4$

Step 15.2.1.2

Multiply $12$ by $1$ .

$f (1) = 12 - 75 {(1)}^{4} + 160 {(1)}^{3} - 120 {(1)}^{2} + 4$

Step 15.2.1.3

One to any power is one.

$f (1) = 12 - 75 \cdot 1 + 160 {(1)}^{3} - 120 {(1)}^{2} + 4$

Step 15.2.1.4

Multiply $- 75$ by $1$ .

$f (1) = 12 - 75 + 160 {(1)}^{3} - 120 {(1)}^{2} + 4$

Step 15.2.1.5

One to any power is one.

$f (1) = 12 - 75 + 160 \cdot 1 - 120 {(1)}^{2} + 4$

Step 15.2.1.6

Multiply $160$ by $1$ .

$f (1) = 12 - 75 + 160 - 120 {(1)}^{2} + 4$

Step 15.2.1.7

One to any power is one.

$f (1) = 12 - 75 + 160 - 120 \cdot 1 + 4$

Step 15.2.1.8

Multiply $- 120$ by $1$ .

$f (1) = 12 - 75 + 160 - 120 + 4$

Step 15.2.2

Simplify by adding and subtracting.

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Step 15.2.2.1

Subtract $75$ from $12$ .

$f (1) = - 63 + 160 - 120 + 4$

Step 15.2.2.2

Add $- 63$ and $160$ .

$f (1) = 97 - 120 + 4$

Step 15.2.2.3

Subtract $120$ from $97$ .

$f (1) = - 23 + 4$

Step 15.2.2.4

Add $- 23$ and $4$ .

$f (1) = - 19$

Step 15.2.3

The final answer is $- 19$ .

$y = - 19$

Step 16

Evaluate the second derivative at $x = 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$240 {(2)}^{3} - 900 {(2)}^{2} + 960 (2) - 240$

Step 17

Evaluate the second derivative.

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Step 17.1

Simplify each term.

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Step 17.1.1

Raise $2$ to the power of $3$ .

$240 \cdot 8 - 900 {(2)}^{2} + 960 (2) - 240$

Step 17.1.2

Multiply $240$ by $8$ .

$1920 - 900 {(2)}^{2} + 960 (2) - 240$

Step 17.1.3

Raise $2$ to the power of $2$ .

$1920 - 900 \cdot 4 + 960 (2) - 240$

Step 17.1.4

Multiply $- 900$ by $4$ .

$1920 - 3600 + 960 (2) - 240$

Step 17.1.5

Multiply $960$ by $2$ .

$1920 - 3600 + 1920 - 240$

Step 17.2

Simplify by adding and subtracting.

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Step 17.2.1

Subtract $3600$ from $1920$ .

$- 1680 + 1920 - 240$

Step 17.2.2

Add $- 1680$ and $1920$ .

$240 - 240$

Step 17.2.3

Subtract $240$ from $240$ .

$0$

Step 18

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 18.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, 0) \cup (0, 1) \cup (1, 2) \cup (2, \infty)$

Step 18.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 0)$ in the first derivative $60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$ to check if the result is negative or positive.

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Step 18.2.1

Replace the variable $x$ with $- 2$ in the expression.

$f' (- 2) = 60 {(- 2)}^{4} - 300 {(- 2)}^{3} + 480 {(- 2)}^{2} - 240 \cdot - 2$

Step 18.2.2

Simplify the result.

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Step 18.2.2.1

Simplify each term.

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Step 18.2.2.1.1

Raise $- 2$ to the power of $4$ .

$f' (- 2) = 60 \cdot 16 - 300 {(- 2)}^{3} + 480 {(- 2)}^{2} - 240 \cdot - 2$

Step 18.2.2.1.2

Multiply $60$ by $16$ .

$f' (- 2) = 960 - 300 {(- 2)}^{3} + 480 {(- 2)}^{2} - 240 \cdot - 2$

Step 18.2.2.1.3

Raise $- 2$ to the power of $3$ .

$f' (- 2) = 960 - 300 \cdot - 8 + 480 {(- 2)}^{2} - 240 \cdot - 2$

Step 18.2.2.1.4

Multiply $- 300$ by $- 8$ .

$f' (- 2) = 960 + 2400 + 480 {(- 2)}^{2} - 240 \cdot - 2$

Step 18.2.2.1.5

Raise $- 2$ to the power of $2$ .

$f' (- 2) = 960 + 2400 + 480 \cdot 4 - 240 \cdot - 2$

Step 18.2.2.1.6

Multiply $480$ by $4$ .

$f' (- 2) = 960 + 2400 + 1920 - 240 \cdot - 2$

Step 18.2.2.1.7

Multiply $- 240$ by $- 2$ .

$f' (- 2) = 960 + 2400 + 1920 + 480$

Step 18.2.2.2

Simplify by adding numbers.

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Step 18.2.2.2.1

Add $960$ and $2400$ .

$f' (- 2) = 3360 + 1920 + 480$

Step 18.2.2.2.2

Add $3360$ and $1920$ .

$f' (- 2) = 5280 + 480$

Step 18.2.2.2.3

Add $5280$ and $480$ .

$f' (- 2) = 5760$

Step 18.2.2.3

The final answer is $5760$ .

$5760$

Step 18.3

Substitute any number, such as $0.5$ , from the interval $(0, 1)$ in the first derivative $60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$ to check if the result is negative or positive.

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Step 18.3.1

Replace the variable $x$ with $0.5$ in the expression.

$f' (0.5) = 60 {(0.5)}^{4} - 300 {(0.5)}^{3} + 480 {(0.5)}^{2} - 240 \cdot 0.5$

Step 18.3.2

Simplify the result.

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Step 18.3.2.1

Simplify each term.

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Step 18.3.2.1.1

Raise $0.5$ to the power of $4$ .

$f' (0.5) = 60 \cdot 0.0625 - 300 {(0.5)}^{3} + 480 {(0.5)}^{2} - 240 \cdot 0.5$

Step 18.3.2.1.2

Multiply $60$ by $0.0625$ .

$f' (0.5) = 3.75 - 300 {(0.5)}^{3} + 480 {(0.5)}^{2} - 240 \cdot 0.5$

Step 18.3.2.1.3

Raise $0.5$ to the power of $3$ .

$f' (0.5) = 3.75 - 300 \cdot 0.125 + 480 {(0.5)}^{2} - 240 \cdot 0.5$

Step 18.3.2.1.4

Multiply $- 300$ by $0.125$ .

$f' (0.5) = 3.75 - 37.5 + 480 {(0.5)}^{2} - 240 \cdot 0.5$

Step 18.3.2.1.5

Raise $0.5$ to the power of $2$ .

$f' (0.5) = 3.75 - 37.5 + 480 \cdot 0.25 - 240 \cdot 0.5$

Step 18.3.2.1.6

Multiply $480$ by $0.25$ .

$f' (0.5) = 3.75 - 37.5 + 120 - 240 \cdot 0.5$

Step 18.3.2.1.7

Multiply $- 240$ by $0.5$ .

$f' (0.5) = 3.75 - 37.5 + 120 - 120$

Step 18.3.2.2

Simplify by adding and subtracting.

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Step 18.3.2.2.1

Subtract $37.5$ from $3.75$ .

$f' (0.5) = - 33.75 + 120 - 120$

Step 18.3.2.2.2

Add $- 33.75$ and $120$ .

$f' (0.5) = 86.25 - 120$

Step 18.3.2.2.3

Subtract $120$ from $86.25$ .

$f' (0.5) = - 33.75$

Step 18.3.2.3

The final answer is $- 33.75$ .

$- 33.75$

Step 18.4

Substitute any number, such as $1.5$ , from the interval $(1, 2)$ in the first derivative $60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$ to check if the result is negative or positive.

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Step 18.4.1

Replace the variable $x$ with $1.5$ in the expression.

$f' (1.5) = 60 {(1.5)}^{4} - 300 {(1.5)}^{3} + 480 {(1.5)}^{2} - 240 \cdot 1.5$

Step 18.4.2

Simplify the result.

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Step 18.4.2.1

Simplify each term.

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Step 18.4.2.1.1

Raise $1.5$ to the power of $4$ .

$f' (1.5) = 60 \cdot 5.0625 - 300 {(1.5)}^{3} + 480 {(1.5)}^{2} - 240 \cdot 1.5$

Step 18.4.2.1.2

Multiply $60$ by $5.0625$ .

$f' (1.5) = 303.75 - 300 {(1.5)}^{3} + 480 {(1.5)}^{2} - 240 \cdot 1.5$

Step 18.4.2.1.3

Raise $1.5$ to the power of $3$ .

$f' (1.5) = 303.75 - 300 \cdot 3.375 + 480 {(1.5)}^{2} - 240 \cdot 1.5$

Step 18.4.2.1.4

Multiply $- 300$ by $3.375$ .

$f' (1.5) = 303.75 - 1012.5 + 480 {(1.5)}^{2} - 240 \cdot 1.5$

Step 18.4.2.1.5

Raise $1.5$ to the power of $2$ .

$f' (1.5) = 303.75 - 1012.5 + 480 \cdot 2.25 - 240 \cdot 1.5$

Step 18.4.2.1.6

Multiply $480$ by $2.25$ .

$f' (1.5) = 303.75 - 1012.5 + 1080 - 240 \cdot 1.5$

Step 18.4.2.1.7

Multiply $- 240$ by $1.5$ .

$f' (1.5) = 303.75 - 1012.5 + 1080 - 360$

Step 18.4.2.2

Simplify by adding and subtracting.

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Step 18.4.2.2.1

Subtract $1012.5$ from $303.75$ .

$f' (1.5) = - 708.75 + 1080 - 360$

Step 18.4.2.2.2

Add $- 708.75$ and $1080$ .

$f' (1.5) = 371.25 - 360$

Step 18.4.2.2.3

Subtract $360$ from $371.25$ .

$f' (1.5) = 11.25$

Step 18.4.2.3

The final answer is $11.25$ .

$11.25$

Step 18.5

Substitute any number, such as $4$ , from the interval $(2, \infty)$ in the first derivative $60 x^{4} - 300 x^{3} + 480 x^{2} - 240 x$ to check if the result is negative or positive.

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Step 18.5.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = 60 {(4)}^{4} - 300 {(4)}^{3} + 480 {(4)}^{2} - 240 \cdot 4$

Step 18.5.2

Simplify the result.

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Step 18.5.2.1

Simplify each term.

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Step 18.5.2.1.1

Raise $4$ to the power of $4$ .

$f' (4) = 60 \cdot 256 - 300 {(4)}^{3} + 480 {(4)}^{2} - 240 \cdot 4$

Step 18.5.2.1.2

Multiply $60$ by $256$ .

$f' (4) = 15360 - 300 {(4)}^{3} + 480 {(4)}^{2} - 240 \cdot 4$

Step 18.5.2.1.3

Raise $4$ to the power of $3$ .

$f' (4) = 15360 - 300 \cdot 64 + 480 {(4)}^{2} - 240 \cdot 4$

Step 18.5.2.1.4

Multiply $- 300$ by $64$ .

$f' (4) = 15360 - 19200 + 480 {(4)}^{2} - 240 \cdot 4$

Step 18.5.2.1.5

Raise $4$ to the power of $2$ .

$f' (4) = 15360 - 19200 + 480 \cdot 16 - 240 \cdot 4$

Step 18.5.2.1.6

Multiply $480$ by $16$ .

$f' (4) = 15360 - 19200 + 7680 - 240 \cdot 4$

Step 18.5.2.1.7

Multiply $- 240$ by $4$ .

$f' (4) = 15360 - 19200 + 7680 - 960$

Step 18.5.2.2

Simplify by adding and subtracting.

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Step 18.5.2.2.1

Subtract $19200$ from $15360$ .

$f' (4) = - 3840 + 7680 - 960$

Step 18.5.2.2.2

Add $- 3840$ and $7680$ .

$f' (4) = 3840 - 960$

Step 18.5.2.2.3

Subtract $960$ from $3840$ .

$f' (4) = 2880$

Step 18.5.2.3

The final answer is $2880$ .

$2880$

Step 18.6

Since the first derivative changed signs from positive to negative around $x = 0$ , then $x = 0$ is a local maximum.

$x = 0$ is a local maximum

Step 18.7

Since the first derivative changed signs from negative to positive around $x = 1$ , then $x = 1$ is a local minimum.

$x = 1$ is a local minimum

Step 18.8

Since the first derivative did not change signs around $x = 2$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 18.9

These are the local extrema for $f (x) = 12 x^{5} - 75 x^{4} + 160 x^{3} - 120 x^{2} + 4$ .

$x = 0$ is a local maximum

$x = 1$ is a local minimum

$x = 0$ is a local maximum

$x = 1$ is a local minimum

Step 19