Calculus Examples

$x^{4} - 12 x^{3} + 48 x^{2} - 64 x$

Step 1

Write $x^{4} - 12 x^{3} + 48 x^{2} - 64 x$ as a function.

$f (x) = x^{4} - 12 x^{3} + 48 x^{2} - 64 x$

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

Differentiate.

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Step 2.1.1.1

By the Sum Rule, the derivative of $x^{4} - 12 x^{3} + 48 x^{2} - 64 x$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 12 x^{3}] + \frac{d}{d x} [48 x^{2}] + \frac{d}{d x} [- 64 x]$ .

$f' (x) = \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 12 x^{3}) + \frac{d}{d x} (48 x^{2}) + \frac{d}{d x} (- 64 x)$

Step 2.1.1.2

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

$f' (x) = 4 x^{3} + \frac{d}{d x} (- 12 x^{3}) + \frac{d}{d x} (48 x^{2}) + \frac{d}{d x} (- 64 x)$

Step 2.1.2

Evaluate $\frac{d}{d x} [- 12 x^{3}]$ .

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Step 2.1.2.1

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 x^{3}$ with respect to $x$ is $- 12 \frac{d}{d x} [x^{3}]$ .

$f' (x) = 4 x^{3} - 12 \frac{d}{d x} x^{3} + \frac{d}{d x} (48 x^{2}) + \frac{d}{d x} (- 64 x)$

Step 2.1.2.2

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

$f' (x) = 4 x^{3} - 12 (3 x^{2}) + \frac{d}{d x} (48 x^{2}) + \frac{d}{d x} (- 64 x)$

Step 2.1.2.3

Multiply $3$ by $- 12$ .

$f' (x) = 4 x^{3} - 36 x^{2} + \frac{d}{d x} (48 x^{2}) + \frac{d}{d x} (- 64 x)$

Step 2.1.3

Evaluate $\frac{d}{d x} [48 x^{2}]$ .

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Step 2.1.3.1

Since $48$ is constant with respect to $x$ , the derivative of $48 x^{2}$ with respect to $x$ is $48 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 4 x^{3} - 36 x^{2} + 48 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 64 x)$

Step 2.1.3.2

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

$f' (x) = 4 x^{3} - 36 x^{2} + 48 (2 x) + \frac{d}{d x} (- 64 x)$

Step 2.1.3.3

Multiply $2$ by $48$ .

$f' (x) = 4 x^{3} - 36 x^{2} + 96 x + \frac{d}{d x} (- 64 x)$

Step 2.1.4

Evaluate $\frac{d}{d x} [- 64 x]$ .

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Step 2.1.4.1

Since $- 64$ is constant with respect to $x$ , the derivative of $- 64 x$ with respect to $x$ is $- 64 \frac{d}{d x} [x]$ .

$f' (x) = 4 x^{3} - 36 x^{2} + 96 x - 64 \frac{d}{d x} x$

Step 2.1.4.2

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

$f' (x) = 4 x^{3} - 36 x^{2} + 96 x - 64 \cdot 1$

Step 2.1.4.3

Multiply $- 64$ by $1$ .

$f' (x) = 4 x^{3} - 36 x^{2} + 96 x - 64$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $4 x^{3} - 36 x^{2} + 96 x - 64$ .

$4 x^{3} - 36 x^{2} + 96 x - 64$

Step 3

Set the first derivative equal to $0$ then solve the equation $4 x^{3} - 36 x^{2} + 96 x - 64 = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$4 x^{3} - 36 x^{2} + 96 x - 64 = 0$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Factor $4$ out of $4 x^{3} - 36 x^{2} + 96 x - 64$ .

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Step 3.2.1.1

Factor $4$ out of $4 x^{3}$ .

$4 (x^{3}) - 36 x^{2} + 96 x - 64 = 0$

Step 3.2.1.2

Factor $4$ out of $- 36 x^{2}$ .

$4 (x^{3}) + 4 (- 9 x^{2}) + 96 x - 64 = 0$

Step 3.2.1.3

Factor $4$ out of $96 x$ .

$4 (x^{3}) + 4 (- 9 x^{2}) + 4 (24 x) - 64 = 0$

Step 3.2.1.4

Factor $4$ out of $- 64$ .

$4 x^{3} + 4 (- 9 x^{2}) + 4 (24 x) + 4 \cdot - 16 = 0$

Step 3.2.1.5

Factor $4$ out of $4 x^{3} + 4 (- 9 x^{2})$ .

$4 (x^{3} - 9 x^{2}) + 4 (24 x) + 4 \cdot - 16 = 0$

Step 3.2.1.6

Factor $4$ out of $4 (x^{3} - 9 x^{2}) + 4 (24 x)$ .

$4 (x^{3} - 9 x^{2} + 24 x) + 4 \cdot - 16 = 0$

Step 3.2.1.7

Factor $4$ out of $4 (x^{3} - 9 x^{2} + 24 x) + 4 \cdot - 16$ .

$4 (x^{3} - 9 x^{2} + 24 x - 16) = 0$

Step 3.2.2

Factor $x^{3} - 9 x^{2} + 24 x - 16$ using the rational roots test.

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Step 3.2.2.1

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.

$p = \pm 1, \pm 16, \pm 2, \pm 8, \pm 4$

$q = \pm 1$

Step 3.2.2.2

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

$\pm 1, \pm 16, \pm 2, \pm 8, \pm 4$

Step 3.2.2.3

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 3.2.2.3.1

Substitute $1$ into the polynomial.

$1^{3} - 9 \cdot 1^{2} + 24 \cdot 1 - 16$

Step 3.2.2.3.2

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

$1 - 9 \cdot 1^{2} + 24 \cdot 1 - 16$

Step 3.2.2.3.3

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

$1 - 9 \cdot 1 + 24 \cdot 1 - 16$

Step 3.2.2.3.4

Multiply $- 9$ by $1$ .

$1 - 9 + 24 \cdot 1 - 16$

Step 3.2.2.3.5

Subtract $9$ from $1$ .

$- 8 + 24 \cdot 1 - 16$

Step 3.2.2.3.6

Multiply $24$ by $1$ .

$- 8 + 24 - 16$

Step 3.2.2.3.7

Add $- 8$ and $24$ .

$16 - 16$

Step 3.2.2.3.8

Subtract $16$ from $16$ .

$0$

Step 3.2.2.4

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.

$\frac{x^{3} - 9 x^{2} + 24 x - 16}{x - 1}$

Step 3.2.2.5

Divide $x^{3} - 9 x^{2} + 24 x - 16$ by $x - 1$ .

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Step 3.2.2.5.1

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

$x$

$1$

$x^{3}$

$9 x^{2}$

$24 x$

$16$

Step 3.2.2.5.2

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$

Step 3.2.2.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			+	$x^{3}$	-	$x^{2}$

Step 3.2.2.5.4

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$

Step 3.2.2.5.5

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$

Step 3.2.2.5.6

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$	+	$24 x$

Step 3.2.2.5.7

Divide the highest order term in the dividend $- 8 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$8 x$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$	+	$24 x$

Step 3.2.2.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$8 x$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$	+	$24 x$
					-	$8 x^{2}$	+	$8 x$

Step 3.2.2.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 8 x^{2} + 8 x$

				$x^{2}$	-	$8 x$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$	+	$24 x$
					+	$8 x^{2}$	-	$8 x$

Step 3.2.2.5.10

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

				$x^{2}$	-	$8 x$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$	+	$24 x$
					+	$8 x^{2}$	-	$8 x$
							+	$16 x$

Step 3.2.2.5.11

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

				$x^{2}$	-	$8 x$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$	+	$24 x$
					+	$8 x^{2}$	-	$8 x$
							+	$16 x$	-	$16$

Step 3.2.2.5.12

Divide the highest order term in the dividend $16 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$	+	$24 x$
					+	$8 x^{2}$	-	$8 x$
							+	$16 x$	-	$16$

Step 3.2.2.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$	+	$24 x$
					+	$8 x^{2}$	-	$8 x$
							+	$16 x$	-	$16$
							+	$16 x$	-	$16$

Step 3.2.2.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $16 x - 16$

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$	+	$24 x$
					+	$8 x^{2}$	-	$8 x$
							+	$16 x$	-	$16$
							-	$16 x$	+	$16$

Step 3.2.2.5.15

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

				$x^{2}$	-	$8 x$	+	$16$
$x$	-	$1$		$x^{3}$	-	$9 x^{2}$	+	$24 x$	-	$16$
			-	$x^{3}$	+	$x^{2}$
					-	$8 x^{2}$	+	$24 x$
					+	$8 x^{2}$	-	$8 x$
							+	$16 x$	-	$16$
							-	$16 x$	+	$16$
										$0$

Step 3.2.2.5.16

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

$x^{2} - 8 x + 16$

Step 3.2.2.6

Write $x^{3} - 9 x^{2} + 24 x - 16$ as a set of factors.

$4 ((x - 1) (x^{2} - 8 x + 16)) = 0$

Step 3.2.3

Factor.

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Step 3.2.3.1

Factor using the perfect square rule.

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Step 3.2.3.1.1

Rewrite $16$ as $4^{2}$ .

$4 ((x - 1) (x^{2} - 8 x + 4^{2})) = 0$

Step 3.2.3.1.2

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

$8 x = 2 \cdot x \cdot 4$

Step 3.2.3.1.3

Rewrite the polynomial.

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

Step 3.2.3.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

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

Step 3.2.3.2

Remove unnecessary parentheses.

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

Step 3.3

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

$x - 1 = 0$

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

Step 3.4

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

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Step 3.4.1

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

$x - 1 = 0$

Step 3.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 3.5

Set ${(x - 4)}^{2}$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set ${(x - 4)}^{2}$ equal to $0$ .

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

Step 3.5.2

Solve ${(x - 4)}^{2} = 0$ for $x$ .

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Step 3.5.2.1

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

$x - 4 = 0$

Step 3.5.2.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.6

The final solution is all the values that make $4 (x - 1) {(x - 4)}^{2} = 0$ true.

$x = 1, 4$

Step 4

The values which make the derivative equal to $0$ are $1, 4$ .

$1, 4$

Step 5

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

$(- \infty, 1) \cup (1, 4) \cup (4, \infty)$

Step 6

Substitute a value from the interval $(- \infty, 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

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

$f' (0) = 4 {(0)}^{3} - 36 {(0)}^{2} + 96 (0) - 64$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

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

$f' (0) = 4 \cdot 0 - 36 {(0)}^{2} + 96 (0) - 64$

Step 6.2.1.2

Multiply $4$ by $0$ .

$f' (0) = 0 - 36 {(0)}^{2} + 96 (0) - 64$

Step 6.2.1.3

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

$f' (0) = 0 - 36 \cdot 0 + 96 (0) - 64$

Step 6.2.1.4

Multiply $- 36$ by $0$ .

$f' (0) = 0 + 0 + 96 (0) - 64$

Step 6.2.1.5

Multiply $96$ by $0$ .

$f' (0) = 0 + 0 + 0 - 64$

Step 6.2.2

Simplify by adding and subtracting.

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Step 6.2.2.1

Add $0$ and $0$ .

$f' (0) = 0 + 0 - 64$

Step 6.2.2.2

Add $0$ and $0$ .

$f' (0) = 0 - 64$

Step 6.2.2.3

Subtract $64$ from $0$ .

$f' (0) = - 64$

Step 6.2.3

The final answer is $- 64$ .

$- 64$

Step 6.3

At $x = 0$ the derivative is $- 64$ . Since this is negative, the function is decreasing on $(- \infty, 1)$ .

Decreasing on $(- \infty, 1)$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(1, 4)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $\frac{5}{2}$ in the expression.

$f' (\frac{5}{2}) = 4 {(\frac{5}{2})}^{3} - 36 {(\frac{5}{2})}^{2} + 96 (\frac{5}{2}) - 64$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Apply the product rule to $\frac{5}{2}$ .

$f' (\frac{5}{2}) = 4 (\frac{5^{3}}{2^{3}}) - 36 {(\frac{5}{2})}^{2} + 96 (\frac{5}{2}) - 64$

Step 7.2.1.2

Raise $5$ to the power of $3$ .

$f' (\frac{5}{2}) = 4 (\frac{125}{2^{3}}) - 36 {(\frac{5}{2})}^{2} + 96 (\frac{5}{2}) - 64$

Step 7.2.1.3

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

$f' (\frac{5}{2}) = 4 (\frac{125}{8}) - 36 {(\frac{5}{2})}^{2} + 96 (\frac{5}{2}) - 64$

Step 7.2.1.4

Cancel the common factor of $4$ .

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Step 7.2.1.4.1

Factor $4$ out of $8$ .

$f' (\frac{5}{2}) = 4 (\frac{125}{4 (2)}) - 36 {(\frac{5}{2})}^{2} + 96 (\frac{5}{2}) - 64$

Step 7.2.1.4.2

Cancel the common factor.

$f' (\frac{5}{2}) = 4 (\frac{125}{4 \cdot 2}) - 36 {(\frac{5}{2})}^{2} + 96 (\frac{5}{2}) - 64$

Step 7.2.1.4.3

Rewrite the expression.

$f' (\frac{5}{2}) = \frac{125}{2} - 36 {(\frac{5}{2})}^{2} + 96 (\frac{5}{2}) - 64$

Step 7.2.1.5

Apply the product rule to $\frac{5}{2}$ .

$f' (\frac{5}{2}) = \frac{125}{2} - 36 \frac{5^{2}}{2^{2}} + 96 (\frac{5}{2}) - 64$

Step 7.2.1.6

Raise $5$ to the power of $2$ .

$f' (\frac{5}{2}) = \frac{125}{2} - 36 \frac{25}{2^{2}} + 96 (\frac{5}{2}) - 64$

Step 7.2.1.7

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

$f' (\frac{5}{2}) = \frac{125}{2} - 36 (\frac{25}{4}) + 96 (\frac{5}{2}) - 64$

Step 7.2.1.8

Cancel the common factor of $4$ .

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Step 7.2.1.8.1

Factor $4$ out of $- 36$ .

$f' (\frac{5}{2}) = \frac{125}{2} + 4 (- 9) (\frac{25}{4}) + 96 (\frac{5}{2}) - 64$

Step 7.2.1.8.2

Cancel the common factor.

$f' (\frac{5}{2}) = \frac{125}{2} + 4 \cdot (- 9 (\frac{25}{4})) + 96 (\frac{5}{2}) - 64$

Step 7.2.1.8.3

Rewrite the expression.

$f' (\frac{5}{2}) = \frac{125}{2} - 9 \cdot 25 + 96 (\frac{5}{2}) - 64$

Step 7.2.1.9

Multiply $- 9$ by $25$ .

$f' (\frac{5}{2}) = \frac{125}{2} - 225 + 96 (\frac{5}{2}) - 64$

Step 7.2.1.10

Cancel the common factor of $2$ .

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Step 7.2.1.10.1

Factor $2$ out of $96$ .

$f' (\frac{5}{2}) = \frac{125}{2} - 225 + 2 (48) (\frac{5}{2}) - 64$

Step 7.2.1.10.2

Cancel the common factor.

$f' (\frac{5}{2}) = \frac{125}{2} - 225 + 2 \cdot (48 (\frac{5}{2})) - 64$

Step 7.2.1.10.3

Rewrite the expression.

$f' (\frac{5}{2}) = \frac{125}{2} - 225 + 48 \cdot 5 - 64$

Step 7.2.1.11

Multiply $48$ by $5$ .

$f' (\frac{5}{2}) = \frac{125}{2} - 225 + 240 - 64$

Step 7.2.2

Find the common denominator.

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Step 7.2.2.1

Write $- 225$ as a fraction with denominator $1$ .

$f' (\frac{5}{2}) = \frac{125}{2} + \frac{- 225}{1} + 240 - 64$

Step 7.2.2.2

Multiply $\frac{- 225}{1}$ by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{125}{2} + \frac{- 225}{1} \cdot \frac{2}{2} + 240 - 64$

Step 7.2.2.3

Multiply $\frac{- 225}{1}$ by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{125}{2} + \frac{- 225 \cdot 2}{2} + 240 - 64$

Step 7.2.2.4

Write $240$ as a fraction with denominator $1$ .

$f' (\frac{5}{2}) = \frac{125}{2} + \frac{- 225 \cdot 2}{2} + \frac{240}{1} - 64$

Step 7.2.2.5

Multiply $\frac{240}{1}$ by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{125}{2} + \frac{- 225 \cdot 2}{2} + \frac{240}{1} \cdot \frac{2}{2} - 64$

Step 7.2.2.6

Multiply $\frac{240}{1}$ by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{125}{2} + \frac{- 225 \cdot 2}{2} + \frac{240 \cdot 2}{2} - 64$

Step 7.2.2.7

Write $- 64$ as a fraction with denominator $1$ .

$f' (\frac{5}{2}) = \frac{125}{2} + \frac{- 225 \cdot 2}{2} + \frac{240 \cdot 2}{2} + \frac{- 64}{1}$

Step 7.2.2.8

Multiply $\frac{- 64}{1}$ by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{125}{2} + \frac{- 225 \cdot 2}{2} + \frac{240 \cdot 2}{2} + \frac{- 64}{1} \cdot \frac{2}{2}$

Step 7.2.2.9

Multiply $\frac{- 64}{1}$ by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{125}{2} + \frac{- 225 \cdot 2}{2} + \frac{240 \cdot 2}{2} + \frac{- 64 \cdot 2}{2}$

Step 7.2.3

Combine the numerators over the common denominator.

$f' (\frac{5}{2}) = \frac{125 - 225 \cdot 2 + 240 \cdot 2 - 64 \cdot 2}{2}$

Step 7.2.4

Simplify each term.

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Step 7.2.4.1

Multiply $- 225$ by $2$ .

$f' (\frac{5}{2}) = \frac{125 - 450 + 240 \cdot 2 - 64 \cdot 2}{2}$

Step 7.2.4.2

Multiply $240$ by $2$ .

$f' (\frac{5}{2}) = \frac{125 - 450 + 480 - 64 \cdot 2}{2}$

Step 7.2.4.3

Multiply $- 64$ by $2$ .

$f' (\frac{5}{2}) = \frac{125 - 450 + 480 - 128}{2}$

Step 7.2.5

Simplify by adding and subtracting.

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Step 7.2.5.1

Subtract $450$ from $125$ .

$f' (\frac{5}{2}) = \frac{- 325 + 480 - 128}{2}$

Step 7.2.5.2

Add $- 325$ and $480$ .

$f' (\frac{5}{2}) = \frac{155 - 128}{2}$

Step 7.2.5.3

Subtract $128$ from $155$ .

$f' (\frac{5}{2}) = \frac{27}{2}$

Step 7.2.6

The final answer is $\frac{27}{2}$ .

$\frac{27}{2}$

Step 7.3

At $x = \frac{5}{2}$ the derivative is $\frac{27}{2}$ . Since this is positive, the function is increasing on $(1, 4)$ .

Increasing on $(1, 4)$ since $f' (x) > 0$

Step 8

Substitute a value from the interval $(4, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $5$ in the expression.

$f' (5) = 4 {(5)}^{3} - 36 {(5)}^{2} + 96 (5) - 64$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Raise $5$ to the power of $3$ .

$f' (5) = 4 \cdot 125 - 36 {(5)}^{2} + 96 (5) - 64$

Step 8.2.1.2

Multiply $4$ by $125$ .

$f' (5) = 500 - 36 {(5)}^{2} + 96 (5) - 64$

Step 8.2.1.3

Raise $5$ to the power of $2$ .

$f' (5) = 500 - 36 \cdot 25 + 96 (5) - 64$

Step 8.2.1.4

Multiply $- 36$ by $25$ .

$f' (5) = 500 - 900 + 96 (5) - 64$

Step 8.2.1.5

Multiply $96$ by $5$ .

$f' (5) = 500 - 900 + 480 - 64$

Step 8.2.2

Simplify by adding and subtracting.

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Step 8.2.2.1

Subtract $900$ from $500$ .

$f' (5) = - 400 + 480 - 64$

Step 8.2.2.2

Add $- 400$ and $480$ .

$f' (5) = 80 - 64$

Step 8.2.2.3

Subtract $64$ from $80$ .

$f' (5) = 16$

Step 8.2.3

The final answer is $16$ .

$16$

Step 8.3

At $x = 5$ the derivative is $16$ . Since this is positive, the function is increasing on $(4, \infty)$ .

Increasing on $(4, \infty)$ since $f' (x) > 0$

Step 9

List the intervals on which the function is increasing and decreasing.

Increasing on: $(1, 4), (4, \infty)$

Decreasing on: $(- \infty, 1)$

Step 10