Calculus Examples

$f (x) = x^{4} - 3 x^{3} + 3 x^{2} - x$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

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

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 3 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- x]$

Step 1.1.2

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

$4 x^{3} + \frac{d}{d x} [- 3 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- x]$

Step 1.2

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

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Step 1.2.1

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

$4 x^{3} - 3 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- x]$

Step 1.2.2

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

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

Step 1.2.3

Multiply $3$ by $- 3$ .

$4 x^{3} - 9 x^{2} + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- x]$

Step 1.3

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

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Step 1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$4 x^{3} - 9 x^{2} + 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x]$

Step 1.3.2

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

$4 x^{3} - 9 x^{2} + 3 (2 x) + \frac{d}{d x} [- x]$

Step 1.3.3

Multiply $2$ by $3$ .

$4 x^{3} - 9 x^{2} + 6 x + \frac{d}{d x} [- x]$

Step 1.4

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

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Step 1.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$4 x^{3} - 9 x^{2} + 6 x - \frac{d}{d x} [x]$

Step 1.4.2

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

$4 x^{3} - 9 x^{2} + 6 x - 1 \cdot 1$

Step 1.4.3

Multiply $- 1$ by $1$ .

$4 x^{3} - 9 x^{2} + 6 x - 1$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $4 x^{3} - 9 x^{2} + 6 x - 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 9 x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 1]$ .

$f'' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (- 9 x^{2}) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 1)$

Step 2.2

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Step 2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 9 x^{2}) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 1)$

Step 2.2.2

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

$f'' (x) = 4 (3 x^{2}) + \frac{d}{d x} (- 9 x^{2}) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 1)$

Step 2.2.3

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (- 9 x^{2}) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 1)$

Step 2.3

Evaluate $\frac{d}{d x} [- 9 x^{2}]$ .

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Step 2.3.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x^{2}$ with respect to $x$ is $- 9 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 12 x^{2} - 9 \frac{d}{d x} x^{2} + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 1)$

Step 2.3.2

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

$f'' (x) = 12 x^{2} - 9 (2 x) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 1)$

Step 2.3.3

Multiply $2$ by $- 9$ .

$f'' (x) = 12 x^{2} - 18 x + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 1)$

Step 2.4

Evaluate $\frac{d}{d x} [6 x]$ .

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Step 2.4.1

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$f'' (x) = 12 x^{2} - 18 x + 6 \frac{d}{d x} (x) + \frac{d}{d x} (- 1)$

Step 2.4.2

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

$f'' (x) = 12 x^{2} - 18 x + 6 \cdot 1 + \frac{d}{d x} (- 1)$

Step 2.4.3

Multiply $6$ by $1$ .

$f'' (x) = 12 x^{2} - 18 x + 6 + \frac{d}{d x} (- 1)$

Step 2.5

Differentiate using the Constant Rule.

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Step 2.5.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f'' (x) = 12 x^{2} - 18 x + 6 + 0$

Step 2.5.2

Add $12 x^{2} - 18 x + 6$ and $0$ .

$f'' (x) = 12 x^{2} - 18 x + 6$

Step 3

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

$4 x^{3} - 9 x^{2} + 6 x - 1 = 0$

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

Differentiate.

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Step 4.1.1.1

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

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 3 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- x]$

Step 4.1.1.2

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

$4 x^{3} + \frac{d}{d x} [- 3 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- x]$

Step 4.1.2

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

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Step 4.1.2.1

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

$4 x^{3} - 3 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- x]$

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 $n = 3$ .

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

Step 4.1.2.3

Multiply $3$ by $- 3$ .

$4 x^{3} - 9 x^{2} + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- x]$

Step 4.1.3

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

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Step 4.1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$4 x^{3} - 9 x^{2} + 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x]$

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 $n = 2$ .

$4 x^{3} - 9 x^{2} + 3 (2 x) + \frac{d}{d x} [- x]$

Step 4.1.3.3

Multiply $2$ by $3$ .

$4 x^{3} - 9 x^{2} + 6 x + \frac{d}{d x} [- x]$

Step 4.1.4

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

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Step 4.1.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$4 x^{3} - 9 x^{2} + 6 x - \frac{d}{d x} [x]$

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 $n = 1$ .

$4 x^{3} - 9 x^{2} + 6 x - 1 \cdot 1$

Step 4.1.4.3

Multiply $- 1$ by $1$ .

$f' (x) = 4 x^{3} - 9 x^{2} + 6 x - 1$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $4 x^{3} - 9 x^{2} + 6 x - 1$ .

$4 x^{3} - 9 x^{2} + 6 x - 1$

Step 5

Set the first derivative equal to $0$ then solve the equation $4 x^{3} - 9 x^{2} + 6 x - 1 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$4 x^{3} - 9 x^{2} + 6 x - 1 = 0$

Step 5.2

Factor the left side of the equation.

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Step 5.2.1

Factor $4 x^{3} - 9 x^{2} + 6 x - 1$ using the rational roots test.

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Step 5.2.1.1

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.

$p = \pm 1$

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

Step 5.2.1.2

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

$\pm 1, \pm 0.25, \pm 0.5$

Step 5.2.1.3

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 5.2.1.3.1

Substitute $1$ into the polynomial.

$4 \cdot 1^{3} - 9 \cdot 1^{2} + 6 \cdot 1 - 1$

Step 5.2.1.3.2

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

$4 \cdot 1 - 9 \cdot 1^{2} + 6 \cdot 1 - 1$

Step 5.2.1.3.3

Multiply $4$ by $1$ .

$4 - 9 \cdot 1^{2} + 6 \cdot 1 - 1$

Step 5.2.1.3.4

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

$4 - 9 \cdot 1 + 6 \cdot 1 - 1$

Step 5.2.1.3.5

Multiply $- 9$ by $1$ .

$4 - 9 + 6 \cdot 1 - 1$

Step 5.2.1.3.6

Subtract $9$ from $4$ .

$- 5 + 6 \cdot 1 - 1$

Step 5.2.1.3.7

Multiply $6$ by $1$ .

$- 5 + 6 - 1$

Step 5.2.1.3.8

Add $- 5$ and $6$ .

$1 - 1$

Step 5.2.1.3.9

Subtract $1$ from $1$ .

$0$

Step 5.2.1.4

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.

$\frac{4 x^{3} - 9 x^{2} + 6 x - 1}{x - 1}$

Step 5.2.1.5

Divide $4 x^{3} - 9 x^{2} + 6 x - 1$ by $x - 1$ .

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Step 5.2.1.5.1

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

$x$

$1$

$4 x^{3}$

$9 x^{2}$

$6 x$

$1$

Step 5.2.1.5.2

Divide the highest order term in the dividend $4 x^{3}$ by the highest order term in divisor $x$ .

					$4 x^{2}$
	$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$

Step 5.2.1.5.3

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			+	$4 x^{3}$	-	$4 x^{2}$

Step 5.2.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $4 x^{3} - 4 x^{2}$

				$4 x^{2}$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$

Step 5.2.1.5.5

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

				$4 x^{2}$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$

Step 5.2.1.5.6

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

				$4 x^{2}$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$	+	$6 x$

Step 5.2.1.5.7

Divide the highest order term in the dividend $- 5 x^{2}$ by the highest order term in divisor $x$ .

				$4 x^{2}$	-	$5 x$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$	+	$6 x$

Step 5.2.1.5.8

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$5 x$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$	+	$6 x$
					-	$5 x^{2}$	+	$5 x$

Step 5.2.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x^{2} + 5 x$

				$4 x^{2}$	-	$5 x$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$	+	$6 x$
					+	$5 x^{2}$	-	$5 x$

Step 5.2.1.5.10

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

				$4 x^{2}$	-	$5 x$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$	+	$6 x$
					+	$5 x^{2}$	-	$5 x$
							+	$x$

Step 5.2.1.5.11

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

				$4 x^{2}$	-	$5 x$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$	+	$6 x$
					+	$5 x^{2}$	-	$5 x$
							+	$x$	-	$1$

Step 5.2.1.5.12

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

				$4 x^{2}$	-	$5 x$	+	$1$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$	+	$6 x$
					+	$5 x^{2}$	-	$5 x$
							+	$x$	-	$1$

Step 5.2.1.5.13

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$5 x$	+	$1$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$	+	$6 x$
					+	$5 x^{2}$	-	$5 x$
							+	$x$	-	$1$
							+	$x$	-	$1$

Step 5.2.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $x - 1$

				$4 x^{2}$	-	$5 x$	+	$1$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$	+	$6 x$
					+	$5 x^{2}$	-	$5 x$
							+	$x$	-	$1$
							-	$x$	+	$1$

Step 5.2.1.5.15

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

				$4 x^{2}$	-	$5 x$	+	$1$
$x$	-	$1$		$4 x^{3}$	-	$9 x^{2}$	+	$6 x$	-	$1$
			-	$4 x^{3}$	+	$4 x^{2}$
					-	$5 x^{2}$	+	$6 x$
					+	$5 x^{2}$	-	$5 x$
							+	$x$	-	$1$
							-	$x$	+	$1$
										$0$

Step 5.2.1.5.16

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

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

Step 5.2.1.6

Write $4 x^{3} - 9 x^{2} + 6 x - 1$ as a set of factors.

$(x - 1) (4 x^{2} - 5 x + 1) = 0$

Step 5.2.2

Factor by grouping.

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Step 5.2.2.1

Factor by grouping.

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Step 5.2.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 1 = 4$ and whose sum is $b = - 5$ .

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Step 5.2.2.1.1.1

Factor $- 5$ out of $- 5 x$ .

$(x - 1) (4 x^{2} - 5 x + 1) = 0$

Step 5.2.2.1.1.2

Rewrite $- 5$ as $- 1$ plus $- 4$

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

Step 5.2.2.1.1.3

Apply the distributive property.

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

Step 5.2.2.1.2

Factor out the greatest common factor from each group.

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Step 5.2.2.1.2.1

Group the first two terms and the last two terms.

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

Step 5.2.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$(x - 1) (x (4 x - 1) - (4 x - 1)) = 0$

Step 5.2.2.1.3

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(x - 1) ((4 x - 1) (x - 1)) = 0$

Step 5.2.2.2

Remove unnecessary parentheses.

$(x - 1) (4 x - 1) (x - 1) = 0$

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 $0$ .

$x - 1 = 0$

$4 x - 1 = 0$

$x - 1 = 0$

Step 5.4

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

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Step 5.4.1

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

$x - 1 = 0$

Step 5.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5.5

Set $4 x - 1$ equal to $0$ and solve for $x$ .

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Step 5.5.1

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Step 5.5.2

Solve $4 x - 1 = 0$ for $x$ .

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Step 5.5.2.1

Add $1$ to both sides of the equation.

$4 x = 1$

Step 5.5.2.2

Divide each term in $4 x = 1$ by $4$ and simplify.

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Step 5.5.2.2.1

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Step 5.5.2.2.2

Simplify the left side.

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Step 5.5.2.2.2.1

Cancel the common factor of $4$ .

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Step 5.5.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Step 5.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 5.6

The final solution is all the values that make $(x - 1) (4 x - 1) (x - 1) = 0$ true.

$x = 1, \frac{1}{4}$

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.

$x = 1, \frac{1}{4}$

Step 8

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.

$12 {(1)}^{2} - 18 \cdot 1 + 6$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

One to any power is one.

$12 \cdot 1 - 18 \cdot 1 + 6$

Step 9.1.2

Multiply $12$ by $1$ .

$12 - 18 \cdot 1 + 6$

Step 9.1.3

Multiply $- 18$ by $1$ .

$12 - 18 + 6$

Step 9.2

Simplify by adding and subtracting.

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Step 9.2.1

Subtract $18$ from $12$ .

$- 6 + 6$

Step 9.2.2

Add $- 6$ and $6$ .

$0$

Step 10

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

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Step 10.1

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

$(- \infty, \frac{1}{4}) \cup (\frac{1}{4}, 1) \cup (1, \infty)$

Step 10.2

Substitute any number, such as $0$ , from the interval $(- \infty, \frac{1}{4})$ in the first derivative $4 x^{3} - 9 x^{2} + 6 x - 1$ to check if the result is negative or positive.

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Step 10.2.1

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

$f' (0) = 4 {(0)}^{3} - 9 {(0)}^{2} + 6 (0) - 1$

Step 10.2.2

Simplify the result.

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Step 10.2.2.1

Simplify each term.

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Step 10.2.2.1.1

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

$f' (0) = 4 \cdot 0 - 9 {(0)}^{2} + 6 (0) - 1$

Step 10.2.2.1.2

Multiply $4$ by $0$ .

$f' (0) = 0 - 9 {(0)}^{2} + 6 (0) - 1$

Step 10.2.2.1.3

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

$f' (0) = 0 - 9 \cdot 0 + 6 (0) - 1$

Step 10.2.2.1.4

Multiply $- 9$ by $0$ .

$f' (0) = 0 + 0 + 6 (0) - 1$

Step 10.2.2.1.5

Multiply $6$ by $0$ .

$f' (0) = 0 + 0 + 0 - 1$

Step 10.2.2.2

Simplify by adding and subtracting.

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Step 10.2.2.2.1

Add $0$ and $0$ .

$f' (0) = 0 + 0 - 1$

Step 10.2.2.2.2

Add $0$ and $0$ .

$f' (0) = 0 - 1$

Step 10.2.2.2.3

Subtract $1$ from $0$ .

$f' (0) = - 1$

Step 10.2.2.3

The final answer is $- 1$ .

$- 1$

Step 10.3

Substitute any number, such as $0.62$ , from the interval $(\frac{1}{4}, 1)$ in the first derivative $4 x^{3} - 9 x^{2} + 6 x - 1$ to check if the result is negative or positive.

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Step 10.3.1

Replace the variable $x$ with $0.62$ in the expression.

$f' (0.62) = 4 {(0.62)}^{3} - 9 {(0.62)}^{2} + 6 (0.62) - 1$

Step 10.3.2

Simplify the result.

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Step 10.3.2.1

Simplify each term.

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Step 10.3.2.1.1

Raise $0.62$ to the power of $3$ .

$f' (0.62) = 4 \cdot 0.238328 - 9 {(0.62)}^{2} + 6 (0.62) - 1$

Step 10.3.2.1.2

Multiply $4$ by $0.238328$ .

$f' (0.62) = 0.953312 - 9 {(0.62)}^{2} + 6 (0.62) - 1$

Step 10.3.2.1.3

Raise $0.62$ to the power of $2$ .

$f' (0.62) = 0.953312 - 9 \cdot 0.3844 + 6 (0.62) - 1$

Step 10.3.2.1.4

Multiply $- 9$ by $0.3844$ .

$f' (0.62) = 0.953312 - 3.4596 + 6 (0.62) - 1$

Step 10.3.2.1.5

Multiply $6$ by $0.62$ .

$f' (0.62) = 0.953312 - 3.4596 + 3.72 - 1$

Step 10.3.2.2

Simplify by adding and subtracting.

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Step 10.3.2.2.1

Subtract $3.4596$ from $0.953312$ .

$f' (0.62) = - 2.506288 + 3.72 - 1$

Step 10.3.2.2.2

Add $- 2.506288$ and $3.72$ .

$f' (0.62) = 1.213712 - 1$

Step 10.3.2.2.3

Subtract $1$ from $1.213712$ .

$f' (0.62) = 0.213712$

Step 10.3.2.3

The final answer is $0.213712$ .

$0.213712$

Step 10.4

Substitute any number, such as $4$ , from the interval $(1, \infty)$ in the first derivative $4 x^{3} - 9 x^{2} + 6 x - 1$ to check if the result is negative or positive.

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Step 10.4.1

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

$f' (4) = 4 {(4)}^{3} - 9 {(4)}^{2} + 6 (4) - 1$

Step 10.4.2

Simplify the result.

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Step 10.4.2.1

Simplify each term.

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Step 10.4.2.1.1

Multiply $4$ by ${(4)}^{3}$ by adding the exponents.

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Step 10.4.2.1.1.1

Multiply $4$ by ${(4)}^{3}$ .

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Step 10.4.2.1.1.1.1

Raise $4$ to the power of $1$ .

$f' (4) = 4 {(4)}^{3} - 9 {(4)}^{2} + 6 (4) - 1$

Step 10.4.2.1.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f' (4) = 4^{1 + 3} - 9 {(4)}^{2} + 6 (4) - 1$

Step 10.4.2.1.1.2

Add $1$ and $3$ .

$f' (4) = 4^{4} - 9 {(4)}^{2} + 6 (4) - 1$

Step 10.4.2.1.2

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

$f' (4) = 256 - 9 {(4)}^{2} + 6 (4) - 1$

Step 10.4.2.1.3

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

$f' (4) = 256 - 9 \cdot 16 + 6 (4) - 1$

Step 10.4.2.1.4

Multiply $- 9$ by $16$ .

$f' (4) = 256 - 144 + 6 (4) - 1$

Step 10.4.2.1.5

Multiply $6$ by $4$ .

$f' (4) = 256 - 144 + 24 - 1$

Step 10.4.2.2

Simplify by adding and subtracting.

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Step 10.4.2.2.1

Subtract $144$ from $256$ .

$f' (4) = 112 + 24 - 1$

Step 10.4.2.2.2

Add $112$ and $24$ .

$f' (4) = 136 - 1$

Step 10.4.2.2.3

Subtract $1$ from $136$ .

$f' (4) = 135$

Step 10.4.2.3

The final answer is $135$ .

$135$

Step 10.5

Since the first derivative changed signs from negative to positive around $x = \frac{1}{4}$ , then $x = \frac{1}{4}$ is a local minimum.

$x = \frac{1}{4}$ is a local minimum

Step 10.6

Since the first derivative did not change signs around $x = 1$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 10.7

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

$x = \frac{1}{4}$ is a local minimum

Step 11