Calculus Examples

$y = \frac{x^{2} - 5}{x - 3}$

Step 1

Write $y = \frac{x^{2} - 5}{x - 3}$ as a function.

$f (x) = \frac{x^{2} - 5}{x - 3}$

Step 2

Find the first derivative of the function.

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Step 2.1

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 $f (x) = x^{2} - 5$ and $g (x) = x - 3$ .

$\frac{(x - 3) \frac{d}{d x} [x^{2} - 5] - (x^{2} - 5) \frac{d}{d x} [x - 3]}{{(x - 3)}^{2}}$

Step 2.2

Differentiate.

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Step 2.2.1

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

$\frac{(x - 3) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 5]) - (x^{2} - 5) \frac{d}{d x} [x - 3]}{{(x - 3)}^{2}}$

Step 2.2.2

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

$\frac{(x - 3) (2 x + \frac{d}{d x} [- 5]) - (x^{2} - 5) \frac{d}{d x} [x - 3]}{{(x - 3)}^{2}}$

Step 2.2.3

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$\frac{(x - 3) (2 x + 0) - (x^{2} - 5) \frac{d}{d x} [x - 3]}{{(x - 3)}^{2}}$

Step 2.2.4

Simplify the expression.

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Step 2.2.4.1

Add $2 x$ and $0$ .

$\frac{(x - 3) (2 x) - (x^{2} - 5) \frac{d}{d x} [x - 3]}{{(x - 3)}^{2}}$

Step 2.2.4.2

Move $2$ to the left of $x - 3$ .

$\frac{2 \cdot (x - 3) x - (x^{2} - 5) \frac{d}{d x} [x - 3]}{{(x - 3)}^{2}}$

Step 2.2.5

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

$\frac{2 (x - 3) x - (x^{2} - 5) (\frac{d}{d x} [x] + \frac{d}{d x} [- 3])}{{(x - 3)}^{2}}$

Step 2.2.6

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

$\frac{2 (x - 3) x - (x^{2} - 5) (1 + \frac{d}{d x} [- 3])}{{(x - 3)}^{2}}$

Step 2.2.7

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$\frac{2 (x - 3) x - (x^{2} - 5) (1 + 0)}{{(x - 3)}^{2}}$

Step 2.2.8

Simplify the expression.

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Step 2.2.8.1

Add $1$ and $0$ .

$\frac{2 (x - 3) x - (x^{2} - 5) \cdot 1}{{(x - 3)}^{2}}$

Step 2.2.8.2

Multiply $- 1$ by $1$ .

$\frac{2 (x - 3) x - (x^{2} - 5)}{{(x - 3)}^{2}}$

Step 2.3

Simplify.

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Step 2.3.1

Apply the distributive property.

$\frac{(2 x + 2 \cdot - 3) x - (x^{2} - 5)}{{(x - 3)}^{2}}$

Step 2.3.2

Apply the distributive property.

$\frac{2 x \cdot x + 2 \cdot - 3 x - (x^{2} - 5)}{{(x - 3)}^{2}}$

Step 2.3.3

Apply the distributive property.

$\frac{2 x \cdot x + 2 \cdot - 3 x - x^{2} - - 5}{{(x - 3)}^{2}}$

Step 2.3.4

Simplify the numerator.

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Step 2.3.4.1

Simplify each term.

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Step 2.3.4.1.1

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

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Step 2.3.4.1.1.1

Move $x$ .

$\frac{2 (x \cdot x) + 2 \cdot - 3 x - x^{2} - - 5}{{(x - 3)}^{2}}$

Step 2.3.4.1.1.2

Multiply $x$ by $x$ .

$\frac{2 x^{2} + 2 \cdot - 3 x - x^{2} - - 5}{{(x - 3)}^{2}}$

Step 2.3.4.1.2

Multiply $2$ by $- 3$ .

$\frac{2 x^{2} - 6 x - x^{2} - - 5}{{(x - 3)}^{2}}$

Step 2.3.4.1.3

Multiply $- 1$ by $- 5$ .

$\frac{2 x^{2} - 6 x - x^{2} + 5}{{(x - 3)}^{2}}$

Step 2.3.4.2

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

$\frac{x^{2} - 6 x + 5}{{(x - 3)}^{2}}$

Step 2.3.5

Factor $x^{2} - 6 x + 5$ using the AC method.

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Step 2.3.5.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $5$ and whose sum is $- 6$ .

$- 5, - 1$

Step 2.3.5.2

Write the factored form using these integers.

$\frac{(x - 5) (x - 1)}{{(x - 3)}^{2}}$

Step 3

Find the second derivative of the function.

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Step 3.1

$f'' (x) = \frac{{(x - 3)}^{2} \frac{d}{d x} ((x - 5) (x - 1)) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{({(x - 3)}^{2})}^{2}}$

Step 3.2

Multiply the exponents in ${({(x - 3)}^{2})}^{2}$ .

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Step 3.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = \frac{{(x - 3)}^{2} \frac{d}{d x} ((x - 5) (x - 1)) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{2 \cdot 2}}$

Step 3.2.2

Multiply $2$ by $2$ .

$f'' (x) = \frac{{(x - 3)}^{2} \frac{d}{d x} ((x - 5) (x - 1)) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x - 5$ and $g (x) = x - 1$ .

$f'' (x) = \frac{{(x - 3)}^{2} ((x - 5) \frac{d}{d x} (x - 1) + (x - 1) \frac{d}{d x} (x - 5)) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4

Differentiate.

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Step 3.4.1

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

$f'' (x) = \frac{{(x - 3)}^{2} ((x - 5) (\frac{d}{d x} (x) + \frac{d}{d x} (- 1)) + (x - 1) \frac{d}{d x} (x - 5)) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.2

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

$f'' (x) = \frac{{(x - 3)}^{2} ((x - 5) (1 + \frac{d}{d x} (- 1)) + (x - 1) \frac{d}{d x} (x - 5)) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.3

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

$f'' (x) = \frac{{(x - 3)}^{2} ((x - 5) (1 + 0) + (x - 1) \frac{d}{d x} (x - 5)) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.4

Simplify the expression.

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Step 3.4.4.1

Add $1$ and $0$ .

$f'' (x) = \frac{{(x - 3)}^{2} ((x - 5) \cdot 1 + (x - 1) \frac{d}{d x} (x - 5)) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.4.2

Multiply $x - 5$ by $1$ .

$f'' (x) = \frac{{(x - 3)}^{2} (x - 5 + (x - 1) \frac{d}{d x} (x - 5)) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.5

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

$f'' (x) = \frac{{(x - 3)}^{2} (x - 5 + (x - 1) (\frac{d}{d x} (x) + \frac{d}{d x} (- 5))) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.6

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

$f'' (x) = \frac{{(x - 3)}^{2} (x - 5 + (x - 1) (1 + \frac{d}{d x} (- 5))) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.7

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$f'' (x) = \frac{{(x - 3)}^{2} (x - 5 + (x - 1) (1 + 0)) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.8

Simplify by adding terms.

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Step 3.4.8.1

Add $1$ and $0$ .

$f'' (x) = \frac{{(x - 3)}^{2} (x - 5 + (x - 1) \cdot 1) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.8.2

Multiply $x - 1$ by $1$ .

$f'' (x) = \frac{{(x - 3)}^{2} (x - 5 + x - 1) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.8.3

Add $x$ and $x$ .

$f'' (x) = \frac{{(x - 3)}^{2} (2 x - 5 - 1) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.4.8.4

Subtract $1$ from $- 5$ .

$f'' (x) = \frac{{(x - 3)}^{2} (2 x - 6) - (x - 5) (x - 1) \frac{d}{d x} {(x - 3)}^{2}}{{(x - 3)}^{4}}$

Step 3.5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = x - 3$ .

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Step 3.5.1

To apply the Chain Rule, set $u$ as $x - 3$ .

$f'' (x) = \frac{{(x - 3)}^{2} (2 x - 6) - (x - 5) (x - 1) (\frac{d}{d u} (u^{2}) \frac{d}{d x} (x - 3))}{{(x - 3)}^{4}}$

Step 3.5.2

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

$f'' (x) = \frac{{(x - 3)}^{2} (2 x - 6) - (x - 5) (x - 1) (2 u \frac{d}{d x} (x - 3))}{{(x - 3)}^{4}}$

Step 3.5.3

Replace all occurrences of $u$ with $x - 3$ .

$f'' (x) = \frac{{(x - 3)}^{2} (2 x - 6) - (x - 5) (x - 1) (2 (x - 3) \frac{d}{d x} (x - 3))}{{(x - 3)}^{4}}$

Step 3.6

Simplify with factoring out.

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Step 3.6.1

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{{(x - 3)}^{2} (2 x - 6) - 2 (x - 5) (x - 1) ((x - 3) \frac{d}{d x} (x - 3))}{{(x - 3)}^{4}}$

Step 3.6.2

Factor $x - 3$ out of ${(x - 3)}^{2} (2 x - 6) - 2 (x - 5) (x - 1) ((x - 3) \frac{d}{d x} [x - 3])$ .

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Step 3.6.2.1

Factor $x - 3$ out of ${(x - 3)}^{2} (2 x - 6)$ .

$f'' (x) = \frac{(x - 3) ((x - 3) (2 x - 6)) - 2 (x - 5) (x - 1) ((x - 3) \frac{d}{d x} (x - 3))}{{(x - 3)}^{4}}$

Step 3.6.2.2

Factor $x - 3$ out of $- 2 (x - 5) (x - 1) ((x - 3) \frac{d}{d x} [x - 3])$ .

$f'' (x) = \frac{(x - 3) ((x - 3) (2 x - 6)) + (x - 3) (- 2 (x - 5) (x - 1) (\frac{d}{d x} (x - 3)))}{{(x - 3)}^{4}}$

Step 3.6.2.3

Factor $x - 3$ out of $(x - 3) ((x - 3) (2 x - 6)) + (x - 3) (- 2 (x - 5) (x - 1) (\frac{d}{d x} [x - 3]))$ .

$f'' (x) = \frac{(x - 3) ((x - 3) (2 x - 6) - 2 (x - 5) (x - 1) (\frac{d}{d x} (x - 3)))}{{(x - 3)}^{4}}$

Step 3.7

Cancel the common factors.

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Step 3.7.1

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

$f'' (x) = \frac{(x - 3) ((x - 3) (2 x - 6) - 2 (x - 5) (x - 1) (\frac{d}{d x} (x - 3)))}{(x - 3) {(x - 3)}^{3}}$

Step 3.7.2

Cancel the common factor.

$f'' (x) = \frac{(x - 3) ((x - 3) (2 x - 6) - 2 (x - 5) (x - 1) (\frac{d}{d x} (x - 3)))}{(x - 3) {(x - 3)}^{3}}$

Step 3.7.3

Rewrite the expression.

$f'' (x) = \frac{(x - 3) (2 x - 6) - 2 (x - 5) (x - 1) (\frac{d}{d x} (x - 3))}{{(x - 3)}^{3}}$

Step 3.8

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

$f'' (x) = \frac{(x - 3) (2 x - 6) - 2 (x - 5) (x - 1) (\frac{d}{d x} (x) + \frac{d}{d x} (- 3))}{{(x - 3)}^{3}}$

Step 3.9

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

$f'' (x) = \frac{(x - 3) (2 x - 6) - 2 (x - 5) (x - 1) (1 + \frac{d}{d x} (- 3))}{{(x - 3)}^{3}}$

Step 3.10

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f'' (x) = \frac{(x - 3) (2 x - 6) - 2 (x - 5) (x - 1) (1 + 0)}{{(x - 3)}^{3}}$

Step 3.11

Simplify the expression.

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Step 3.11.1

Add $1$ and $0$ .

$f'' (x) = \frac{(x - 3) (2 x - 6) - 2 (x - 5) (x - 1) \cdot 1}{{(x - 3)}^{3}}$

Step 3.11.2

Multiply $- 2$ by $1$ .

$f'' (x) = \frac{(x - 3) (2 x - 6) - 2 (x - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12

Simplify.

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Step 3.12.1

Apply the distributive property.

$f'' (x) = \frac{(x - 3) (2 x - 6) + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2

Simplify the numerator.

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Step 3.12.2.1

Simplify each term.

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Step 3.12.2.1.1

Expand $(x - 3) (2 x - 6)$ using the FOIL Method.

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Step 3.12.2.1.1.1

Apply the distributive property.

$f'' (x) = \frac{x (2 x - 6) - 3 (2 x - 6) + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.1.2

Apply the distributive property.

$f'' (x) = \frac{x (2 x) + x \cdot - 6 - 3 (2 x - 6) + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.1.3

Apply the distributive property.

$f'' (x) = \frac{x (2 x) + x \cdot - 6 - 3 (2 x) - 3 \cdot - 6 + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.2

Simplify and combine like terms.

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Step 3.12.2.1.2.1

Simplify each term.

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Step 3.12.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{2 x \cdot x + x \cdot - 6 - 3 (2 x) - 3 \cdot - 6 + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.2.1.2

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

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Step 3.12.2.1.2.1.2.1

Move $x$ .

$f'' (x) = \frac{2 (x \cdot x) + x \cdot - 6 - 3 (2 x) - 3 \cdot - 6 + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.2.1.2.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{2 x^{2} + x \cdot - 6 - 3 (2 x) - 3 \cdot - 6 + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.2.1.3

Move $- 6$ to the left of $x$ .

$f'' (x) = \frac{2 x^{2} - 6 \cdot x - 3 (2 x) - 3 \cdot - 6 + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.2.1.4

Multiply $2$ by $- 3$ .

$f'' (x) = \frac{2 x^{2} - 6 x - 6 x - 3 \cdot - 6 + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.2.1.5

Multiply $- 3$ by $- 6$ .

$f'' (x) = \frac{2 x^{2} - 6 x - 6 x + 18 + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.2.2

Subtract $6 x$ from $- 6 x$ .

$f'' (x) = \frac{2 x^{2} - 12 x + 18 + (- 2 x - 2 \cdot - 5) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.3

Multiply $- 2$ by $- 5$ .

$f'' (x) = \frac{2 x^{2} - 12 x + 18 + (- 2 x + 10) (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.4

Expand $(- 2 x + 10) (x - 1)$ using the FOIL Method.

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Step 3.12.2.1.4.1

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} - 12 x + 18 - 2 x (x - 1) + 10 (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.4.2

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} - 12 x + 18 - 2 x \cdot x - 2 x \cdot - 1 + 10 (x - 1)}{{(x - 3)}^{3}}$

Step 3.12.2.1.4.3

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} - 12 x + 18 - 2 x \cdot x - 2 x \cdot - 1 + 10 x + 10 \cdot - 1}{{(x - 3)}^{3}}$

Step 3.12.2.1.5

Simplify and combine like terms.

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Step 3.12.2.1.5.1

Simplify each term.

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Step 3.12.2.1.5.1.1

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

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Step 3.12.2.1.5.1.1.1

Move $x$ .

$f'' (x) = \frac{2 x^{2} - 12 x + 18 - 2 (x \cdot x) - 2 x \cdot - 1 + 10 x + 10 \cdot - 1}{{(x - 3)}^{3}}$

Step 3.12.2.1.5.1.1.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{2 x^{2} - 12 x + 18 - 2 x^{2} - 2 x \cdot - 1 + 10 x + 10 \cdot - 1}{{(x - 3)}^{3}}$

Step 3.12.2.1.5.1.2

Multiply $- 1$ by $- 2$ .

$f'' (x) = \frac{2 x^{2} - 12 x + 18 - 2 x^{2} + 2 x + 10 x + 10 \cdot - 1}{{(x - 3)}^{3}}$

Step 3.12.2.1.5.1.3

Multiply $10$ by $- 1$ .

$f'' (x) = \frac{2 x^{2} - 12 x + 18 - 2 x^{2} + 2 x + 10 x - 10}{{(x - 3)}^{3}}$

Step 3.12.2.1.5.2

Add $2 x$ and $10 x$ .

$f'' (x) = \frac{2 x^{2} - 12 x + 18 - 2 x^{2} + 12 x - 10}{{(x - 3)}^{3}}$

Step 3.12.2.2

Combine the opposite terms in $2 x^{2} - 12 x + 18 - 2 x^{2} + 12 x - 10$ .

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Step 3.12.2.2.1

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

$f'' (x) = \frac{- 12 x + 18 + 0 + 12 x - 10}{{(x - 3)}^{3}}$

Step 3.12.2.2.2

Add $- 12 x + 18$ and $0$ .

$f'' (x) = \frac{- 12 x + 18 + 12 x - 10}{{(x - 3)}^{3}}$

Step 3.12.2.2.3

Add $- 12 x$ and $12 x$ .

$f'' (x) = \frac{0 + 18 - 10}{{(x - 3)}^{3}}$

Step 3.12.2.2.4

Add $0$ and $18$ .

$f'' (x) = \frac{18 - 10}{{(x - 3)}^{3}}$

Step 3.12.2.3

Subtract $10$ from $18$ .

$f'' (x) = \frac{8}{{(x - 3)}^{3}}$

Step 4

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

$\frac{(x - 5) (x - 1)}{{(x - 3)}^{2}} = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

$f' (x) = \frac{(x - 3) \frac{d}{d x} (x^{2} - 5) - (x^{2} - 5) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 5.1.2

Differentiate.

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Step 5.1.2.1

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

$f' (x) = \frac{(x - 3) (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 5)) - (x^{2} - 5) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 5.1.2.2

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

$f' (x) = \frac{(x - 3) (2 x + \frac{d}{d x} (- 5)) - (x^{2} - 5) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 5.1.2.3

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$f' (x) = \frac{(x - 3) (2 x + 0) - (x^{2} - 5) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 5.1.2.4

Simplify the expression.

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Step 5.1.2.4.1

Add $2 x$ and $0$ .

$f' (x) = \frac{(x - 3) (2 x) - (x^{2} - 5) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 5.1.2.4.2

Move $2$ to the left of $x - 3$ .

$f' (x) = \frac{2 \cdot ((x - 3) x) - (x^{2} - 5) \frac{d}{d x} (x - 3)}{{(x - 3)}^{2}}$

Step 5.1.2.5

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

$f' (x) = \frac{2 (x - 3) x - (x^{2} - 5) (\frac{d}{d x} (x) + \frac{d}{d x} (- 3))}{{(x - 3)}^{2}}$

Step 5.1.2.6

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

$f' (x) = \frac{2 (x - 3) x - (x^{2} - 5) (1 + \frac{d}{d x} (- 3))}{{(x - 3)}^{2}}$

Step 5.1.2.7

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f' (x) = \frac{2 (x - 3) x - (x^{2} - 5) (1 + 0)}{{(x - 3)}^{2}}$

Step 5.1.2.8

Simplify the expression.

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Step 5.1.2.8.1

Add $1$ and $0$ .

$f' (x) = \frac{2 (x - 3) x - (x^{2} - 5) \cdot 1}{{(x - 3)}^{2}}$

Step 5.1.2.8.2

Multiply $- 1$ by $1$ .

$f' (x) = \frac{2 (x - 3) x - (x^{2} - 5)}{{(x - 3)}^{2}}$

Step 5.1.3

Simplify.

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Step 5.1.3.1

Apply the distributive property.

$f' (x) = \frac{(2 x + 2 \cdot - 3) x - (x^{2} - 5)}{{(x - 3)}^{2}}$

Step 5.1.3.2

Apply the distributive property.

$f' (x) = \frac{2 x \cdot x + 2 \cdot (- 3 x) - (x^{2} - 5)}{{(x - 3)}^{2}}$

Step 5.1.3.3

Apply the distributive property.

$f' (x) = \frac{2 x \cdot x + 2 \cdot (- 3 x) - x^{2} + 5}{{(x - 3)}^{2}}$

Step 5.1.3.4

Simplify the numerator.

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Step 5.1.3.4.1

Simplify each term.

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Step 5.1.3.4.1.1

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

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Step 5.1.3.4.1.1.1

Move $x$ .

$f' (x) = \frac{2 (x \cdot x) + 2 \cdot (- 3 x) - x^{2} + 5}{{(x - 3)}^{2}}$

Step 5.1.3.4.1.1.2

Multiply $x$ by $x$ .

$f' (x) = \frac{2 x^{2} + 2 \cdot (- 3 x) - x^{2} + 5}{{(x - 3)}^{2}}$

Step 5.1.3.4.1.2

Multiply $2$ by $- 3$ .

$f' (x) = \frac{2 x^{2} - 6 x - x^{2} + 5}{{(x - 3)}^{2}}$

Step 5.1.3.4.1.3

Multiply $- 1$ by $- 5$ .

$f' (x) = \frac{2 x^{2} - 6 x - x^{2} + 5}{{(x - 3)}^{2}}$

Step 5.1.3.4.2

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

$f' (x) = \frac{x^{2} - 6 x + 5}{{(x - 3)}^{2}}$

Step 5.1.3.5

Factor $x^{2} - 6 x + 5$ using the AC method.

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Step 5.1.3.5.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $5$ and whose sum is $- 6$ .

$- 5, - 1$

Step 5.1.3.5.2

Write the factored form using these integers.

$f' (x) = \frac{(x - 5) (x - 1)}{{(x - 3)}^{2}}$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{(x - 5) (x - 1)}{{(x - 3)}^{2}}$ .

$\frac{(x - 5) (x - 1)}{{(x - 3)}^{2}}$

Step 6

Set the first derivative equal to $0$ then solve the equation $\frac{(x - 5) (x - 1)}{{(x - 3)}^{2}} = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$\frac{(x - 5) (x - 1)}{{(x - 3)}^{2}} = 0$

Step 6.2

Set the numerator equal to zero.

$(x - 5) (x - 1) = 0$

Step 6.3

Solve the equation for $x$ .

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Step 6.3.1

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

$x - 5 = 0$

$x - 1 = 0$

Step 6.3.2

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 6.3.2.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 6.3.2.2

Add $5$ to both sides of the equation.

$x = 5$

Step 6.3.3

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

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Step 6.3.3.1

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

$x - 1 = 0$

Step 6.3.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 6.3.4

The final solution is all the values that make $(x - 5) (x - 1) = 0$ true.

$x = 5, 1$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

Set the denominator in $\frac{(x - 5) (x - 1)}{{(x - 3)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x - 3)}^{2} = 0$

Step 7.2

Solve for $x$ .

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Step 7.2.1

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 7.2.2

Add $3$ to both sides of the equation.

$x = 3$

Step 8

Critical points to evaluate.

$x = 5, 1$

Step 9

Evaluate the second derivative at $x = 5$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{8}{{((5) - 3)}^{3}}$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify the denominator.

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Step 10.1.1

Subtract $3$ from $5$ .

$\frac{8}{2^{3}}$

Step 10.1.2

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

$\frac{8}{8}$

Step 10.2

Divide $8$ by $8$ .

$1$

Step 11

$x = 5$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 5$ is a local minimum

Step 12

Find the y-value when $x = 5$ .

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Step 12.1

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

$f (5) = \frac{{(5)}^{2} - 5}{(5) - 3}$

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify the numerator.

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Step 12.2.1.1

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

$f (5) = \frac{25 - 5}{5 - 3}$

Step 12.2.1.2

Subtract $5$ from $25$ .

$f (5) = \frac{20}{5 - 3}$

Step 12.2.2

Simplify the expression.

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Step 12.2.2.1

Subtract $3$ from $5$ .

$f (5) = \frac{20}{2}$

Step 12.2.2.2

Divide $20$ by $2$ .

$f (5) = 10$

Step 12.2.3

The final answer is $10$ .

$y = 10$

Step 13

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.

$\frac{8}{{((1) - 3)}^{3}}$

Step 14

Evaluate the second derivative.

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Step 14.1

Simplify the denominator.

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Step 14.1.1

Subtract $3$ from $1$ .

$\frac{8}{{(- 2)}^{3}}$

Step 14.1.2

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

$\frac{8}{- 8}$

Step 14.2

Divide $8$ by $- 8$ .

$- 1$

Step 15

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

$x = 1$ is a local maximum

Step 16

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

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Step 16.1

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

$f (1) = \frac{{(1)}^{2} - 5}{(1) - 3}$

Step 16.2

Simplify the result.

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Step 16.2.1

Simplify the numerator.

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Step 16.2.1.1

One to any power is one.

$f (1) = \frac{1 - 5}{1 - 3}$

Step 16.2.1.2

Subtract $5$ from $1$ .

$f (1) = \frac{- 4}{1 - 3}$

Step 16.2.2

Simplify the expression.

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Step 16.2.2.1

Subtract $3$ from $1$ .

$f (1) = \frac{- 4}{- 2}$

Step 16.2.2.2

Divide $- 4$ by $- 2$ .

$f (1) = 2$

Step 16.2.3

The final answer is $2$ .

$y = 2$

Step 17

These are the local extrema for $f (x) = \frac{x^{2} - 5}{x - 3}$ .

$(5, 10)$ is a local minima

$(1, 2)$ is a local maxima

Step 18