Calculus Examples

$y = x^{4} - 3 x^{3} + 3 x^{2} - x$

Step 1

Write $y = x^{4} - 3 x^{3} + 3 x^{2} - x$ as a function.

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

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 $x^{4} - 3 x^{3} + 3 x^{2} - x$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 3 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- x]$ .

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

Step 2.1.1.2

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

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

Step 2.1.2

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

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Step 2.1.2.1

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

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

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 $n = 3$ .

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

Step 2.1.2.3

Multiply $3$ by $- 3$ .

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

Step 2.1.3

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

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Step 2.1.3.1

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

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

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 $n = 2$ .

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

Step 2.1.3.3

Multiply $2$ by $3$ .

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

Step 2.1.4

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

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Step 2.1.4.1

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

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

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 $n = 1$ .

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

Step 2.1.4.3

Multiply $- 1$ by $1$ .

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

Step 2.2

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 3

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 3.1

Set the first derivative equal to $0$ .

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

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

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

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Step 3.2.1.1

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 3.2.1.2

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 3.2.1.3

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 3.2.1.3.1

Substitute $1$ into the polynomial.

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

Step 3.2.1.3.2

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

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

Step 3.2.1.3.3

Multiply $4$ by $1$ .

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

Step 3.2.1.3.4

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

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

Step 3.2.1.3.5

Multiply $- 9$ by $1$ .

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

Step 3.2.1.3.6

Subtract $9$ from $4$ .

$- 5 + 6 \cdot 1 - 1$

Step 3.2.1.3.7

Multiply $6$ by $1$ .

$- 5 + 6 - 1$

Step 3.2.1.3.8

Add $- 5$ and $6$ .

$1 - 1$

Step 3.2.1.3.9

Subtract $1$ from $1$ .

$0$

Step 3.2.1.4

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 3.2.1.5

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

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Step 3.2.1.5.1

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 3.2.1.5.2

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 3.2.1.5.3

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 3.2.1.5.4

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 3.2.1.5.5

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 3.2.1.5.6

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 3.2.1.5.7

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 3.2.1.5.8

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 3.2.1.5.9

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 3.2.1.5.10

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 3.2.1.5.11

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 3.2.1.5.12

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 3.2.1.5.13

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 3.2.1.5.14

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 3.2.1.5.15

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 3.2.1.5.16

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

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

Step 3.2.1.6

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 3.2.2

Factor by grouping.

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Step 3.2.2.1

Factor by grouping.

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Step 3.2.2.1.1

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 3.2.2.1.1.1

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

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

Step 3.2.2.1.1.2

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

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

Step 3.2.2.1.1.3

Apply the distributive property.

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

Step 3.2.2.1.2

Factor out the greatest common factor from each group.

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Step 3.2.2.1.2.1

Group the first two terms and the last two terms.

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

Step 3.2.2.1.2.2

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

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

Step 3.2.2.1.3

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

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

Step 3.2.2.2

Remove unnecessary parentheses.

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

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 $0$ .

$x - 1 = 0$

$4 x - 1 = 0$

$x - 1 = 0$

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

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Step 3.5.1

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

$4 x - 1 = 0$

Step 3.5.2

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

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Step 3.5.2.1

Add $1$ to both sides of the equation.

$4 x = 1$

Step 3.5.2.2

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

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Step 3.5.2.2.1

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

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

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Cancel the common factor of $4$ .

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Step 3.5.2.2.2.1.1

Cancel the common factor.

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

Step 3.5.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 3.6

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

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

Step 4

The values which make the derivative equal to $0$ are $1, \frac{1}{4}$ .

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

Step 5

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

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

Step 6

Substitute a value from the interval $(- \infty, \frac{1}{4})$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 0.75$ in the expression.

$f' (- 0.75) = 4 {(- 0.75)}^{3} - 9 {(- 0.75)}^{2} + 6 (- 0.75) - 1$

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

Raise $- 0.75$ to the power of $3$ .

$f' (- 0.75) = 4 \cdot - 0.421875 - 9 {(- 0.75)}^{2} + 6 (- 0.75) - 1$

Step 6.2.1.2

Multiply $4$ by $- 0.421875$ .

$f' (- 0.75) = - 1.6875 - 9 {(- 0.75)}^{2} + 6 (- 0.75) - 1$

Step 6.2.1.3

Raise $- 0.75$ to the power of $2$ .

$f' (- 0.75) = - 1.6875 - 9 \cdot 0.5625 + 6 (- 0.75) - 1$

Step 6.2.1.4

Multiply $- 9$ by $0.5625$ .

$f' (- 0.75) = - 1.6875 - 5.0625 + 6 (- 0.75) - 1$

Step 6.2.1.5

Multiply $6$ by $- 0.75$ .

$f' (- 0.75) = - 1.6875 - 5.0625 - 4.5 - 1$

Step 6.2.2

Simplify by subtracting numbers.

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Step 6.2.2.1

Subtract $5.0625$ from $- 1.6875$ .

$f' (- 0.75) = - 6.75 - 4.5 - 1$

Step 6.2.2.2

Subtract $4.5$ from $- 6.75$ .

$f' (- 0.75) = - 11.25 - 1$

Step 6.2.2.3

Subtract $1$ from $- 11.25$ .

$f' (- 0.75) = - 12.25$

Step 6.2.3

The final answer is $- 12.25$ .

$- 12.25$

Step 6.3

At $x = - 0.75$ the derivative is $- 12.25$ . Since this is negative, the function is decreasing on $(- \infty, \frac{1}{4})$ .

Decreasing on $(- \infty, \frac{1}{4})$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(0.25, 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.625$ in the expression.

$f' (0.625) = 4 {(0.625)}^{3} - 9 {(0.625)}^{2} + 6 (0.625) - 1$

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

Raise $0.625$ to the power of $3$ .

$f' (0.625) = 4 \cdot 0.24414062 - 9 {(0.625)}^{2} + 6 (0.625) - 1$

Step 7.2.1.2

Multiply $4$ by $0.24414062$ .

$f' (0.625) = 0.9765625 - 9 {(0.625)}^{2} + 6 (0.625) - 1$

Step 7.2.1.3

Raise $0.625$ to the power of $2$ .

$f' (0.625) = 0.9765625 - 9 \cdot 0.390625 + 6 (0.625) - 1$

Step 7.2.1.4

Multiply $- 9$ by $0.390625$ .

$f' (0.625) = 0.9765625 - 3.515625 + 6 (0.625) - 1$

Step 7.2.1.5

Multiply $6$ by $0.625$ .

$f' (0.625) = 0.9765625 - 3.515625 + 3.75 - 1$

Step 7.2.2

Simplify by adding and subtracting.

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Step 7.2.2.1

Subtract $3.515625$ from $0.9765625$ .

$f' (0.625) = - 2.5390625 + 3.75 - 1$

Step 7.2.2.2

Add $- 2.5390625$ and $3.75$ .

$f' (0.625) = 1.2109375 - 1$

Step 7.2.2.3

Subtract $1$ from $1.2109375$ .

$f' (0.625) = 0.2109375$

Step 7.2.3

The final answer is $0.2109375$ .

$0.2109375$

Step 7.3

At $x = 0.625$ the derivative is $0.2109375$ . Since this is positive, the function is increasing on $(0.25, 1)$ .

Increasing on $(\frac{1}{4}, 1)$ since $f' (x) > 0$

Step 8

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

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Step 8.1

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

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

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 $2$ to the power of $3$ .

$f' (2) = 4 \cdot 8 - 9 {(2)}^{2} + 6 (2) - 1$

Step 8.2.1.2

Multiply $4$ by $8$ .

$f' (2) = 32 - 9 {(2)}^{2} + 6 (2) - 1$

Step 8.2.1.3

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

$f' (2) = 32 - 9 \cdot 4 + 6 (2) - 1$

Step 8.2.1.4

Multiply $- 9$ by $4$ .

$f' (2) = 32 - 36 + 6 (2) - 1$

Step 8.2.1.5

Multiply $6$ by $2$ .

$f' (2) = 32 - 36 + 12 - 1$

Step 8.2.2

Simplify by adding and subtracting.

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Step 8.2.2.1

Subtract $36$ from $32$ .

$f' (2) = - 4 + 12 - 1$

Step 8.2.2.2

Add $- 4$ and $12$ .

$f' (2) = 8 - 1$

Step 8.2.2.3

Subtract $1$ from $8$ .

$f' (2) = 7$

Step 8.2.3

The final answer is $7$ .

$7$

Step 8.3

At $x = 2$ the derivative is $7$ . Since this is positive, the function is increasing on $(1, \infty)$ .

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

Step 9

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

Increasing on: $(\frac{1}{4}, 1), (1, \infty)$

Decreasing on: $(- \infty, \frac{1}{4})$

Step 10