Calculus Examples

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

Step 1

Find the first derivative of the function.

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

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

Move $2$ to the left of $x - 5$ .

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.2.5

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

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

Step 1.2.6

Simplify the expression.

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Step 1.2.6.1

Add $1$ and $0$ .

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

Step 1.2.6.2

Multiply $- 1$ by $1$ .

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

Step 1.3

Simplify.

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Step 1.3.1

Apply the distributive property.

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

Step 1.3.2

Apply the distributive property.

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

Step 1.3.3

Simplify the numerator.

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Step 1.3.3.1

Simplify each term.

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Step 1.3.3.1.1

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

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Step 1.3.3.1.1.1

Move $x$ .

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

Step 1.3.3.1.1.2

Multiply $x$ by $x$ .

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

Step 1.3.3.1.2

Multiply $2$ by $- 5$ .

$\frac{2 x^{2} - 10 x - x^{2}}{{(x - 5)}^{2}}$

Step 1.3.3.2

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

$\frac{x^{2} - 10 x}{{(x - 5)}^{2}}$

Step 1.3.4

Factor $x$ out of $x^{2} - 10 x$ .

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Step 1.3.4.1

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

$\frac{x \cdot x - 10 x}{{(x - 5)}^{2}}$

Step 1.3.4.2

Factor $x$ out of $- 10 x$ .

$\frac{x \cdot x + x \cdot - 10}{{(x - 5)}^{2}}$

Step 1.3.4.3

Factor $x$ out of $x \cdot x + x \cdot - 10$ .

$\frac{x (x - 10)}{{(x - 5)}^{2}}$

Step 2

Find the second derivative of the function.

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Step 2.1

$f'' (x) = \frac{{(x - 5)}^{2} \frac{d}{d x} (x (x - 10)) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{({(x - 5)}^{2})}^{2}}$

Step 2.2

Multiply the exponents in ${({(x - 5)}^{2})}^{2}$ .

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Step 2.2.1

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

$f'' (x) = \frac{{(x - 5)}^{2} \frac{d}{d x} (x (x - 10)) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{2 \cdot 2}}$

Step 2.2.2

Multiply $2$ by $2$ .

$f'' (x) = \frac{{(x - 5)}^{2} \frac{d}{d x} (x (x - 10)) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{4}}$

Step 2.3

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 $f (x) = x$ and $g (x) = x - 10$ .

$f'' (x) = \frac{{(x - 5)}^{2} (x \frac{d}{d x} (x - 10) + (x - 10) \frac{d}{d x} (x)) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{4}}$

Step 2.4

Differentiate.

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Step 2.4.1

By the Sum Rule, the derivative of $x - 10$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 10]$ .

$f'' (x) = \frac{{(x - 5)}^{2} (x (\frac{d}{d x} (x) + \frac{d}{d x} (- 10)) + (x - 10) \frac{d}{d x} (x)) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{4}}$

Step 2.4.2

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

$f'' (x) = \frac{{(x - 5)}^{2} (x (1 + \frac{d}{d x} (- 10)) + (x - 10) \frac{d}{d x} (x)) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{4}}$

Step 2.4.3

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10$ with respect to $x$ is $0$ .

$f'' (x) = \frac{{(x - 5)}^{2} (x (1 + 0) + (x - 10) \frac{d}{d x} (x)) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{4}}$

Step 2.4.4

Simplify the expression.

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Step 2.4.4.1

Add $1$ and $0$ .

$f'' (x) = \frac{{(x - 5)}^{2} (x \cdot 1 + (x - 10) \frac{d}{d x} (x)) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{4}}$

Step 2.4.4.2

Multiply $x$ by $1$ .

$f'' (x) = \frac{{(x - 5)}^{2} (x + (x - 10) \frac{d}{d x} (x)) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{4}}$

Step 2.4.5

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

$f'' (x) = \frac{{(x - 5)}^{2} (x + (x - 10) \cdot 1) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{4}}$

Step 2.4.6

Simplify by adding terms.

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Step 2.4.6.1

Multiply $x - 10$ by $1$ .

$f'' (x) = \frac{{(x - 5)}^{2} (x + x - 10) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{4}}$

Step 2.4.6.2

Add $x$ and $x$ .

$f'' (x) = \frac{{(x - 5)}^{2} (2 x - 10) - x (x - 10) \frac{d}{d x} {(x - 5)}^{2}}{{(x - 5)}^{4}}$

Step 2.5

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 $g (x) = x - 5$ .

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Step 2.5.1

To apply the Chain Rule, set $u$ as $x - 5$ .

$f'' (x) = \frac{{(x - 5)}^{2} (2 x - 10) - x (x - 10) (\frac{d}{d u} (u^{2}) \frac{d}{d x} (x - 5))}{{(x - 5)}^{4}}$

Step 2.5.2

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

$f'' (x) = \frac{{(x - 5)}^{2} (2 x - 10) - x (x - 10) (2 u \frac{d}{d x} (x - 5))}{{(x - 5)}^{4}}$

Step 2.5.3

Replace all occurrences of $u$ with $x - 5$ .

$f'' (x) = \frac{{(x - 5)}^{2} (2 x - 10) - x (x - 10) (2 (x - 5) \frac{d}{d x} (x - 5))}{{(x - 5)}^{4}}$

Step 2.6

Simplify with factoring out.

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Step 2.6.1

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{{(x - 5)}^{2} (2 x - 10) - 2 x (x - 10) ((x - 5) \frac{d}{d x} (x - 5))}{{(x - 5)}^{4}}$

Step 2.6.2

Factor $x - 5$ out of ${(x - 5)}^{2} (2 x - 10) - 2 x (x - 10) ((x - 5) \frac{d}{d x} [x - 5])$ .

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Step 2.6.2.1

Factor $x - 5$ out of ${(x - 5)}^{2} (2 x - 10)$ .

$f'' (x) = \frac{(x - 5) ((x - 5) (2 x - 10)) - 2 x (x - 10) ((x - 5) \frac{d}{d x} (x - 5))}{{(x - 5)}^{4}}$

Step 2.6.2.2

Factor $x - 5$ out of $- 2 x (x - 10) ((x - 5) \frac{d}{d x} [x - 5])$ .

$f'' (x) = \frac{(x - 5) ((x - 5) (2 x - 10)) + (x - 5) (- 2 x (x - 10) (\frac{d}{d x} (x - 5)))}{{(x - 5)}^{4}}$

Step 2.6.2.3

Factor $x - 5$ out of $(x - 5) ((x - 5) (2 x - 10)) + (x - 5) (- 2 x (x - 10) (\frac{d}{d x} [x - 5]))$ .

$f'' (x) = \frac{(x - 5) ((x - 5) (2 x - 10) - 2 x (x - 10) (\frac{d}{d x} (x - 5)))}{{(x - 5)}^{4}}$

Step 2.7

Cancel the common factors.

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Step 2.7.1

Factor $x - 5$ out of ${(x - 5)}^{4}$ .

$f'' (x) = \frac{(x - 5) ((x - 5) (2 x - 10) - 2 x (x - 10) (\frac{d}{d x} (x - 5)))}{(x - 5) {(x - 5)}^{3}}$

Step 2.7.2

Cancel the common factor.

$f'' (x) = \frac{(x - 5) ((x - 5) (2 x - 10) - 2 x (x - 10) (\frac{d}{d x} (x - 5)))}{(x - 5) {(x - 5)}^{3}}$

Step 2.7.3

Rewrite the expression.

$f'' (x) = \frac{(x - 5) (2 x - 10) - 2 x (x - 10) (\frac{d}{d x} (x - 5))}{{(x - 5)}^{3}}$

Step 2.8

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

$f'' (x) = \frac{(x - 5) (2 x - 10) - 2 x (x - 10) (\frac{d}{d x} (x) + \frac{d}{d x} (- 5))}{{(x - 5)}^{3}}$

Step 2.9

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

$f'' (x) = \frac{(x - 5) (2 x - 10) - 2 x (x - 10) (1 + \frac{d}{d x} (- 5))}{{(x - 5)}^{3}}$

Step 2.10

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

$f'' (x) = \frac{(x - 5) (2 x - 10) - 2 x (x - 10) (1 + 0)}{{(x - 5)}^{3}}$

Step 2.11

Simplify the expression.

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Step 2.11.1

Add $1$ and $0$ .

$f'' (x) = \frac{(x - 5) (2 x - 10) - 2 x (x - 10) \cdot 1}{{(x - 5)}^{3}}$

Step 2.11.2

Multiply $- 2$ by $1$ .

$f'' (x) = \frac{(x - 5) (2 x - 10) - 2 x (x - 10)}{{(x - 5)}^{3}}$

Step 2.12

Simplify.

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Step 2.12.1

Apply the distributive property.

$f'' (x) = \frac{(x - 5) (2 x - 10) - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2

Simplify the numerator.

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Step 2.12.2.1

Simplify each term.

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Step 2.12.2.1.1

Expand $(x - 5) (2 x - 10)$ using the FOIL Method.

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Step 2.12.2.1.1.1

Apply the distributive property.

$f'' (x) = \frac{x (2 x - 10) - 5 (2 x - 10) - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.1.2

Apply the distributive property.

$f'' (x) = \frac{x (2 x) + x \cdot - 10 - 5 (2 x - 10) - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.1.3

Apply the distributive property.

$f'' (x) = \frac{x (2 x) + x \cdot - 10 - 5 (2 x) - 5 \cdot - 10 - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.2

Simplify and combine like terms.

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Step 2.12.2.1.2.1

Simplify each term.

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Step 2.12.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{2 x \cdot x + x \cdot - 10 - 5 (2 x) - 5 \cdot - 10 - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.2.1.2

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

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Step 2.12.2.1.2.1.2.1

Move $x$ .

$f'' (x) = \frac{2 (x \cdot x) + x \cdot - 10 - 5 (2 x) - 5 \cdot - 10 - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.2.1.2.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{2 x^{2} + x \cdot - 10 - 5 (2 x) - 5 \cdot - 10 - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.2.1.3

Move $- 10$ to the left of $x$ .

$f'' (x) = \frac{2 x^{2} - 10 \cdot x - 5 (2 x) - 5 \cdot - 10 - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.2.1.4

Multiply $2$ by $- 5$ .

$f'' (x) = \frac{2 x^{2} - 10 x - 10 x - 5 \cdot - 10 - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.2.1.5

Multiply $- 5$ by $- 10$ .

$f'' (x) = \frac{2 x^{2} - 10 x - 10 x + 50 - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.2.2

Subtract $10 x$ from $- 10 x$ .

$f'' (x) = \frac{2 x^{2} - 20 x + 50 - 2 x \cdot x - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.3

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

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Step 2.12.2.1.3.1

Move $x$ .

$f'' (x) = \frac{2 x^{2} - 20 x + 50 - 2 (x \cdot x) - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.3.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{2 x^{2} - 20 x + 50 - 2 x^{2} - 2 x \cdot - 10}{{(x - 5)}^{3}}$

Step 2.12.2.1.4

Multiply $- 10$ by $- 2$ .

$f'' (x) = \frac{2 x^{2} - 20 x + 50 - 2 x^{2} + 20 x}{{(x - 5)}^{3}}$

Step 2.12.2.2

Combine the opposite terms in $2 x^{2} - 20 x + 50 - 2 x^{2} + 20 x$ .

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Step 2.12.2.2.1

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

$f'' (x) = \frac{- 20 x + 50 + 0 + 20 x}{{(x - 5)}^{3}}$

Step 2.12.2.2.2

Add $- 20 x + 50$ and $0$ .

$f'' (x) = \frac{- 20 x + 50 + 20 x}{{(x - 5)}^{3}}$

Step 2.12.2.2.3

Add $- 20 x$ and $20 x$ .

$f'' (x) = \frac{0 + 50}{{(x - 5)}^{3}}$

Step 2.12.2.2.4

Add $0$ and $50$ .

$f'' (x) = \frac{50}{{(x - 5)}^{3}}$

Step 3

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

$\frac{x (x - 10)}{{(x - 5)}^{2}} = 0$

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

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

Step 4.1.2

Differentiate.

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Step 4.1.2.1

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

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

Step 4.1.2.2

Move $2$ to the left of $x - 5$ .

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

Step 4.1.2.3

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

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

Step 4.1.2.4

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

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

Step 4.1.2.5

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

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

Step 4.1.2.6

Simplify the expression.

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Step 4.1.2.6.1

Add $1$ and $0$ .

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

Step 4.1.2.6.2

Multiply $- 1$ by $1$ .

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

Step 4.1.3

Simplify.

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Step 4.1.3.1

Apply the distributive property.

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

Step 4.1.3.2

Apply the distributive property.

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

Step 4.1.3.3

Simplify the numerator.

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Step 4.1.3.3.1

Simplify each term.

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Step 4.1.3.3.1.1

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

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Step 4.1.3.3.1.1.1

Move $x$ .

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

Step 4.1.3.3.1.1.2

Multiply $x$ by $x$ .

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

Step 4.1.3.3.1.2

Multiply $2$ by $- 5$ .

$\frac{2 x^{2} - 10 x - x^{2}}{{(x - 5)}^{2}}$

Step 4.1.3.3.2

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

$\frac{x^{2} - 10 x}{{(x - 5)}^{2}}$

Step 4.1.3.4

Factor $x$ out of $x^{2} - 10 x$ .

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Step 4.1.3.4.1

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

$\frac{x \cdot x - 10 x}{{(x - 5)}^{2}}$

Step 4.1.3.4.2

Factor $x$ out of $- 10 x$ .

$\frac{x \cdot x + x \cdot - 10}{{(x - 5)}^{2}}$

Step 4.1.3.4.3

Factor $x$ out of $x \cdot x + x \cdot - 10$ .

$f' (x) = \frac{x (x - 10)}{{(x - 5)}^{2}}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{x (x - 10)}{{(x - 5)}^{2}}$ .

$\frac{x (x - 10)}{{(x - 5)}^{2}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{x (x - 10)}{{(x - 5)}^{2}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{x (x - 10)}{{(x - 5)}^{2}} = 0$

Step 5.2

Set the numerator equal to zero.

$x (x - 10) = 0$

Step 5.3

Solve the equation for $x$ .

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Step 5.3.1

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

$x = 0$

$x - 10 = 0$

Step 5.3.2

Set $x$ equal to $0$ .

$x = 0$

Step 5.3.3

Set $x - 10$ equal to $0$ and solve for $x$ .

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Step 5.3.3.1

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Step 5.3.3.2

Add $10$ to both sides of the equation.

$x = 10$

Step 5.3.4

The final solution is all the values that make $x (x - 10) = 0$ true.

$x = 0, 10$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{x (x - 10)}{{(x - 5)}^{2}}$ equal to $0$ to find where the expression is undefined.

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

Step 6.2

Solve for $x$ .

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Step 6.2.1

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

$x - 5 = 0$

Step 6.2.2

Add $5$ to both sides of the equation.

$x = 5$

Step 7

Critical points to evaluate.

$x = 0, 10$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{50}{{((0) - 5)}^{3}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the denominator.

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Step 9.1.1

Subtract $5$ from $0$ .

$\frac{50}{{(- 5)}^{3}}$

Step 9.1.2

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

$\frac{50}{- 125}$

Step 9.2

Reduce the expression by cancelling the common factors.

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Step 9.2.1

Cancel the common factor of $50$ and $- 125$ .

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Step 9.2.1.1

Factor $25$ out of $50$ .

$\frac{25 (2)}{- 125}$

Step 9.2.1.2

Cancel the common factors.

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Step 9.2.1.2.1

Factor $25$ out of $- 125$ .

$\frac{25 \cdot 2}{25 \cdot - 5}$

Step 9.2.1.2.2

Cancel the common factor.

$\frac{25 \cdot 2}{25 \cdot - 5}$

Step 9.2.1.2.3

Rewrite the expression.

$\frac{2}{- 5}$

Step 9.2.2

Move the negative in front of the fraction.

$- \frac{2}{5}$

Step 10

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 11

Find the y-value when $x = 0$ .

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Step 11.1

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

$f (0) = \frac{{(0)}^{2}}{(0) - 5}$

Step 11.2

Simplify the result.

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Step 11.2.1

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

$f (0) = \frac{0}{0 - 5}$

Step 11.2.2

Subtract $5$ from $0$ .

$f (0) = \frac{0}{- 5}$

Step 11.2.3

Divide $0$ by $- 5$ .

$f (0) = 0$

Step 11.2.4

The final answer is $0$ .

$y = 0$

Step 12

Evaluate the second derivative at $x = 10$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{50}{{((10) - 5)}^{3}}$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify the denominator.

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Step 13.1.1

Subtract $5$ from $10$ .

$\frac{50}{5^{3}}$

Step 13.1.2

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

$\frac{50}{125}$

Step 13.2

Cancel the common factor of $50$ and $125$ .

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Step 13.2.1

Factor $25$ out of $50$ .

$\frac{25 (2)}{125}$

Step 13.2.2

Cancel the common factors.

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Step 13.2.2.1

Factor $25$ out of $125$ .

$\frac{25 \cdot 2}{25 \cdot 5}$

Step 13.2.2.2

Cancel the common factor.

$\frac{25 \cdot 2}{25 \cdot 5}$

Step 13.2.2.3

Rewrite the expression.

$\frac{2}{5}$

Step 14

$x = 10$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 10$ is a local minimum

Step 15

Find the y-value when $x = 10$ .

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Step 15.1

Replace the variable $x$ with $10$ in the expression.

$f (10) = \frac{{(10)}^{2}}{(10) - 5}$

Step 15.2

Simplify the result.

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Step 15.2.1

Raise $10$ to the power of $2$ .

$f (10) = \frac{100}{10 - 5}$

Step 15.2.2

Subtract $5$ from $10$ .

$f (10) = \frac{100}{5}$

Step 15.2.3

Divide $100$ by $5$ .

$f (10) = 20$

Step 15.2.4

The final answer is $20$ .

$y = 20$

Step 16

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

$(0, 0)$ is a local maxima

$(10, 20)$ is a local minima

Step 17