Calculus Examples

$f (x) = \frac{x^{2} - 7 x + 12}{x^{2} - 16}$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x^{2} - 7 x + 12$ and $g (x) = x^{2} - 16$ .

$\frac{(x^{2} - 16) \frac{d}{d x} [x^{2} - 7 x + 12] - (x^{2} - 7 x + 12) \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $x^{2} - 7 x + 12$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 7 x] + \frac{d}{d x} [12]$ .

$\frac{(x^{2} - 16) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 7 x] + \frac{d}{d x} [12]) - (x^{2} - 7 x + 12) \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.2

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

$\frac{(x^{2} - 16) (2 x + \frac{d}{d x} [- 7 x] + \frac{d}{d x} [12]) - (x^{2} - 7 x + 12) \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.3

Since $- 7$ is constant with respect to $x$ , the derivative of $- 7 x$ with respect to $x$ is $- 7 \frac{d}{d x} [x]$ .

$\frac{(x^{2} - 16) (2 x - 7 \frac{d}{d x} [x] + \frac{d}{d x} [12]) - (x^{2} - 7 x + 12) \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.4

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

$\frac{(x^{2} - 16) (2 x - 7 \cdot 1 + \frac{d}{d x} [12]) - (x^{2} - 7 x + 12) \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.5

Multiply $- 7$ by $1$ .

$\frac{(x^{2} - 16) (2 x - 7 + \frac{d}{d x} [12]) - (x^{2} - 7 x + 12) \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.6

Since $12$ is constant with respect to $x$ , the derivative of $12$ with respect to $x$ is $0$ .

$\frac{(x^{2} - 16) (2 x - 7 + 0) - (x^{2} - 7 x + 12) \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.7

Add $2 x - 7$ and $0$ .

$\frac{(x^{2} - 16) (2 x - 7) - (x^{2} - 7 x + 12) \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.8

By the Sum Rule, the derivative of $x^{2} - 16$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]$ .

$\frac{(x^{2} - 16) (2 x - 7) - (x^{2} - 7 x + 12) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16])}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.9

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

$\frac{(x^{2} - 16) (2 x - 7) - (x^{2} - 7 x + 12) (2 x + \frac{d}{d x} [- 16])}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.10

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16$ with respect to $x$ is $0$ .

$\frac{(x^{2} - 16) (2 x - 7) - (x^{2} - 7 x + 12) (2 x + 0)}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.11

Simplify the expression.

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Step 1.1.2.11.1

Add $2 x$ and $0$ .

$\frac{(x^{2} - 16) (2 x - 7) - (x^{2} - 7 x + 12) (2 x)}{{(x^{2} - 16)}^{2}}$

Step 1.1.2.11.2

Multiply $2$ by $- 1$ .

$\frac{(x^{2} - 16) (2 x - 7) - 2 (x^{2} - 7 x + 12) x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3

Simplify.

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Step 1.1.3.1

Apply the distributive property.

$\frac{(x^{2} - 16) (2 x - 7) + (- 2 x^{2} - 2 (- 7 x) - 2 \cdot 12) x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.2

Apply the distributive property.

$\frac{(x^{2} - 16) (2 x - 7) - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3

Simplify the numerator.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

Expand $(x^{2} - 16) (2 x - 7)$ using the FOIL Method.

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Step 1.1.3.3.1.1.1

Apply the distributive property.

$\frac{x^{2} (2 x - 7) - 16 (2 x - 7) - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.1.2

Apply the distributive property.

$\frac{x^{2} (2 x) + x^{2} \cdot - 7 - 16 (2 x - 7) - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.1.3

Apply the distributive property.

$\frac{x^{2} (2 x) + x^{2} \cdot - 7 - 16 (2 x) - 16 \cdot - 7 - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.2

Simplify each term.

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Step 1.1.3.3.1.2.1

Rewrite using the commutative property of multiplication.

$\frac{2 x^{2} x + x^{2} \cdot - 7 - 16 (2 x) - 16 \cdot - 7 - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.2.2

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.3.3.1.2.2.1

Move $x$ .

$\frac{2 (x \cdot x^{2}) + x^{2} \cdot - 7 - 16 (2 x) - 16 \cdot - 7 - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.2.2.2

Multiply $x$ by $x^{2}$ .

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Step 1.1.3.3.1.2.2.2.1

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

$\frac{2 (x^{1} x^{2}) + x^{2} \cdot - 7 - 16 (2 x) - 16 \cdot - 7 - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.2.2.2.2

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

$\frac{2 x^{1 + 2} + x^{2} \cdot - 7 - 16 (2 x) - 16 \cdot - 7 - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.2.2.3

Add $1$ and $2$ .

$\frac{2 x^{3} + x^{2} \cdot - 7 - 16 (2 x) - 16 \cdot - 7 - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.2.3

Move $- 7$ to the left of $x^{2}$ .

$\frac{2 x^{3} - 7 \cdot x^{2} - 16 (2 x) - 16 \cdot - 7 - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.2.4

Multiply $2$ by $- 16$ .

$\frac{2 x^{3} - 7 x^{2} - 32 x - 16 \cdot - 7 - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.2.5

Multiply $- 16$ by $- 7$ .

$\frac{2 x^{3} - 7 x^{2} - 32 x + 112 - 2 x^{2} x - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.3

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.3.3.1.3.1

Move $x$ .

$\frac{2 x^{3} - 7 x^{2} - 32 x + 112 - 2 (x \cdot x^{2}) - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.3.2

Multiply $x$ by $x^{2}$ .

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Step 1.1.3.3.1.3.2.1

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

$\frac{2 x^{3} - 7 x^{2} - 32 x + 112 - 2 (x^{1} x^{2}) - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.3.2.2

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

$\frac{2 x^{3} - 7 x^{2} - 32 x + 112 - 2 x^{1 + 2} - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.3.3

Add $1$ and $2$ .

$\frac{2 x^{3} - 7 x^{2} - 32 x + 112 - 2 x^{3} - 2 (- 7 x) x - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.4

Multiply $x$ by $x$ by adding the exponents.

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Step 1.1.3.3.1.4.1

Move $x$ .

$\frac{2 x^{3} - 7 x^{2} - 32 x + 112 - 2 x^{3} - 2 (- 7 (x \cdot x)) - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.4.2

Multiply $x$ by $x$ .

$\frac{2 x^{3} - 7 x^{2} - 32 x + 112 - 2 x^{3} - 2 (- 7 x^{2}) - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.5

Multiply $- 7$ by $- 2$ .

$\frac{2 x^{3} - 7 x^{2} - 32 x + 112 - 2 x^{3} + 14 x^{2} - 2 \cdot 12 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.1.6

Multiply $- 2$ by $12$ .

$\frac{2 x^{3} - 7 x^{2} - 32 x + 112 - 2 x^{3} + 14 x^{2} - 24 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.2

Combine the opposite terms in $2 x^{3} - 7 x^{2} - 32 x + 112 - 2 x^{3} + 14 x^{2} - 24 x$ .

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Step 1.1.3.3.2.1

Subtract $2 x^{3}$ from $2 x^{3}$ .

$\frac{- 7 x^{2} - 32 x + 112 + 0 + 14 x^{2} - 24 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.2.2

Add $- 7 x^{2} - 32 x + 112$ and $0$ .

$\frac{- 7 x^{2} - 32 x + 112 + 14 x^{2} - 24 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.3

Add $- 7 x^{2}$ and $14 x^{2}$ .

$\frac{7 x^{2} - 32 x + 112 - 24 x}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.3.4

Subtract $24 x$ from $- 32 x$ .

$\frac{7 x^{2} - 56 x + 112}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.4

Simplify the numerator.

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Step 1.1.3.4.1

Factor $7$ out of $7 x^{2} - 56 x + 112$ .

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Step 1.1.3.4.1.1

Factor $7$ out of $7 x^{2}$ .

$\frac{7 (x^{2}) - 56 x + 112}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.4.1.2

Factor $7$ out of $- 56 x$ .

$\frac{7 (x^{2}) + 7 (- 8 x) + 112}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.4.1.3

Factor $7$ out of $112$ .

$\frac{7 x^{2} + 7 (- 8 x) + 7 \cdot 16}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.4.1.4

Factor $7$ out of $7 x^{2} + 7 (- 8 x)$ .

$\frac{7 (x^{2} - 8 x) + 7 \cdot 16}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.4.1.5

Factor $7$ out of $7 (x^{2} - 8 x) + 7 \cdot 16$ .

$\frac{7 (x^{2} - 8 x + 16)}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.4.2

Factor using the perfect square rule.

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Step 1.1.3.4.2.1

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

$\frac{7 (x^{2} - 8 x + 4^{2})}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.4.2.2

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 1.1.3.4.2.3

Rewrite the polynomial.

$\frac{7 (x^{2} - 2 \cdot x \cdot 4 + 4^{2})}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.4.2.4

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

$\frac{7 {(x - 4)}^{2}}{{(x^{2} - 16)}^{2}}$

Step 1.1.3.5

Simplify the denominator.

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Step 1.1.3.5.1

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

$\frac{7 {(x - 4)}^{2}}{{(x^{2} - 4^{2})}^{2}}$

Step 1.1.3.5.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

$\frac{7 {(x - 4)}^{2}}{{((x + 4) (x - 4))}^{2}}$

Step 1.1.3.5.3

Apply the product rule to $(x + 4) (x - 4)$ .

$\frac{7 {(x - 4)}^{2}}{{(x + 4)}^{2} {(x - 4)}^{2}}$

Step 1.1.3.6

Cancel the common factor of ${(x - 4)}^{2}$ .

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Step 1.1.3.6.1

Cancel the common factor.

$\frac{7 {(x - 4)}^{2}}{{(x + 4)}^{2} {(x - 4)}^{2}}$

Step 1.1.3.6.2

Rewrite the expression.

$f' (x) = \frac{7}{{(x + 4)}^{2}}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{7}{{(x + 4)}^{2}}$ .

$\frac{7}{{(x + 4)}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{7}{{(x + 4)}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$\frac{7}{{(x + 4)}^{2}} = 0$

Step 2.2

Set the numerator equal to zero.

$7 = 0$

Step 2.3

Since $7 \neq 0$ , there are no solutions.

No solution

Step 3

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{7}{{(x + 4)}^{2}}$ equal to $0$ to find where the expression is undefined.

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

Step 4.2

Solve for $x$ .

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Step 4.2.1

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

$x + 4 = 0$

Step 4.2.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5

After finding the point that makes the derivative $f' (x) = \frac{7}{{(x + 4)}^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{x^{2} - 7 x + 12}{x^{2} - 16}$ is increasing and where it is decreasing is $(- \infty, - 4) \cup (- 4, \infty)$ .

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

Step 6

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

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Step 6.1

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

$f' (- 5) = \frac{7}{{((- 5) + 4)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the denominator.

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Step 6.2.1.1

Add $- 5$ and $4$ .

$f' (- 5) = \frac{7}{{(- 1)}^{2}}$

Step 6.2.1.2

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

$f' (- 5) = \frac{7}{1}$

Step 6.2.2

Divide $7$ by $1$ .

$f' (- 5) = 7$

Step 6.2.3

The final answer is $7$ .

$7$

Step 6.3

At $x = - 5$ the derivative is $7$ . Since this is positive, the function is increasing on $(- \infty, - 4)$ .

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

Step 7

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 7.1

Replace the variable $x$ with $- 3$ in the expression.

$f' (- 3) = \frac{7}{{((- 3) + 4)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the denominator.

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Step 7.2.1.1

Add $- 3$ and $4$ .

$f' (- 3) = \frac{7}{1^{2}}$

Step 7.2.1.2

One to any power is one.

$f' (- 3) = \frac{7}{1}$

Step 7.2.2

Divide $7$ by $1$ .

$f' (- 3) = 7$

Step 7.2.3

The final answer is $7$ .

$7$

Step 7.3

At $x = - 3$ the derivative is $7$ . Since this is positive, the function is increasing on $(- 4, \infty)$ .

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

Step 8

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

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

Step 9