Calculus Examples

$f (x) = \frac{1 + x^{2}}{1 - x^{2}}$

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 $f (x) = 1 + x^{2}$ and $g (x) = 1 - x^{2}$ .

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

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $1 + x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [x^{2}]$ .

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Add $0$ and $\frac{d}{d x} [x^{2}]$ .

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

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 $n = 2$ .

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

Step 1.1.2.5

Move $2$ to the left of $1 - x^{2}$ .

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

Step 1.1.2.6

By the Sum Rule, the derivative of $1 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- x^{2}]$ .

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

Step 1.1.2.7

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

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

Step 1.1.2.8

Add $0$ and $\frac{d}{d x} [- x^{2}]$ .

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

Step 1.1.2.9

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Step 1.1.2.10

Multiply.

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Step 1.1.2.10.1

Multiply $- 1$ by $- 1$ .

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

Step 1.1.2.10.2

Multiply $1 + x^{2}$ by $1$ .

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

Step 1.1.2.11

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

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

Step 1.1.2.12

Move $2$ to the left of $1 + x^{2}$ .

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

Step 1.1.3

Simplify.

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Step 1.1.3.1

Apply the distributive property.

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

Step 1.1.3.2

Apply the distributive property.

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

Step 1.1.3.3

Apply the distributive property.

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

Step 1.1.3.4

Apply the distributive property.

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

Step 1.1.3.5

Simplify the numerator.

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Step 1.1.3.5.1

Simplify each term.

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Step 1.1.3.5.1.1

Multiply $2$ by $1$ .

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

Step 1.1.3.5.1.2

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

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Step 1.1.3.5.1.2.1

Move $x$ .

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

Step 1.1.3.5.1.2.2

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

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Step 1.1.3.5.1.2.2.1

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

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

Step 1.1.3.5.1.2.2.2

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

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

Step 1.1.3.5.1.2.3

Add $1$ and $2$ .

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

Step 1.1.3.5.1.3

Multiply $- 1$ by $2$ .

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

Step 1.1.3.5.1.4

Multiply $2$ by $1$ .

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

Step 1.1.3.5.1.5

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

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Step 1.1.3.5.1.5.1

Move $x$ .

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

Step 1.1.3.5.1.5.2

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

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Step 1.1.3.5.1.5.2.1

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

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

Step 1.1.3.5.1.5.2.2

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

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

Step 1.1.3.5.1.5.3

Add $1$ and $2$ .

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

Step 1.1.3.5.2

Combine the opposite terms in $2 x - 2 x^{3} + 2 x + 2 x^{3}$ .

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Step 1.1.3.5.2.1

Add $- 2 x^{3}$ and $2 x^{3}$ .

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

Step 1.1.3.5.2.2

Add $2 x + 2 x$ and $0$ .

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

Step 1.1.3.5.3

Add $2 x$ and $2 x$ .

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

Step 1.1.3.6

Reorder terms.

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

Step 1.1.3.7

Simplify the denominator.

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Step 1.1.3.7.1

Rewrite $1$ as $1^{2}$ .

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

Step 1.1.3.7.2

Reorder $- x^{2}$ and $1^{2}$ .

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

Step 1.1.3.7.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

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

Step 1.1.3.7.4

Apply the product rule to $(1 + x) (1 - x)$ .

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

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{4 x}{{(1 + x)}^{2} {(1 - x)}^{2}}$ .

$\frac{4 x}{{(1 + x)}^{2} {(1 - x)}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{4 x}{{(1 + x)}^{2} {(1 - x)}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

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

Step 2.2

Set the numerator equal to zero.

$4 x = 0$

Step 2.3

Divide each term in $4 x = 0$ by $4$ and simplify.

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Step 2.3.1

Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $4$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $0$ by $4$ .

$x = 0$

Step 3

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

$0$

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{4 x}{{(1 + x)}^{2} {(1 - x)}^{2}}$ equal to $0$ to find where the expression is undefined.

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

Step 4.2

Solve for $x$ .

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Step 4.2.1

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

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

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

Step 4.2.2

Set ${(1 + x)}^{2}$ equal to $0$ and solve for $x$ .

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Step 4.2.2.1

Set ${(1 + x)}^{2}$ equal to $0$ .

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

Step 4.2.2.2

Solve ${(1 + x)}^{2} = 0$ for $x$ .

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Step 4.2.2.2.1

Set the $1 + x$ equal to $0$ .

$1 + x = 0$

Step 4.2.2.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.2.3

Set ${(1 - x)}^{2}$ equal to $0$ and solve for $x$ .

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Step 4.2.3.1

Set ${(1 - x)}^{2}$ equal to $0$ .

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

Step 4.2.3.2

Solve ${(1 - x)}^{2} = 0$ for $x$ .

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Step 4.2.3.2.1

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

$1 - x = 0$

Step 4.2.3.2.2

Solve for $x$ .

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Step 4.2.3.2.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 4.2.3.2.2.2

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

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Step 4.2.3.2.2.2.1

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

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

Step 4.2.3.2.2.2.2

Simplify the left side.

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Step 4.2.3.2.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 4.2.3.2.2.2.2.2

Divide $x$ by $1$ .

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

Step 4.2.3.2.2.2.3

Simplify the right side.

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Step 4.2.3.2.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 4.2.4

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

$x = - 1, 1$

Step 4.3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 1, x = 1$

Step 5

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

$(- \infty, - 1) \cup (- 1, 0) \cup (0, 1) \cup (1, \infty)$

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 $- 2$ in the expression.

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

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the expression.

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Step 6.2.1.1

Remove parentheses.

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

Step 6.2.1.2

Multiply $4$ by $- 2$ .

$f' (- 2) = \frac{- 8}{{(1 - 2)}^{2} {(1 - (- 2))}^{2}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Subtract $2$ from $1$ .

$f' (- 2) = \frac{- 8}{{(- 1)}^{2} {(1 - (- 2))}^{2}}$

Step 6.2.2.2

Multiply $- 1$ by $- 2$ .

$f' (- 2) = \frac{- 8}{{(- 1)}^{2} {(1 + 2)}^{2}}$

Step 6.2.2.3

Add $1$ and $2$ .

$f' (- 2) = \frac{- 8}{{(- 1)}^{2} \cdot 3^{2}}$

Step 6.2.2.4

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

$f' (- 2) = \frac{- 8}{1 \cdot 3^{2}}$

Step 6.2.2.5

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

$f' (- 2) = \frac{- 8}{1 \cdot 9}$

Step 6.2.2.6

Multiply $9$ by $1$ .

$f' (- 2) = \frac{- 8}{9}$

Step 6.2.3

Move the negative in front of the fraction.

$f' (- 2) = - \frac{8}{9}$

Step 6.2.4

The final answer is $- \frac{8}{9}$ .

$- \frac{8}{9}$

Step 6.3

At $x = - 2$ the derivative is $- \frac{8}{9}$ . 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 $(- 1, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

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

Step 7.2

Simplify the result.

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Step 7.2.1

Remove parentheses.

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

Step 7.2.2

Simplify the numerator.

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Step 7.2.2.1

Multiply $- 1$ by $4$ .

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

Step 7.2.2.2

Combine $- 4$ and $\frac{1}{2}$ .

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

Step 7.2.3

Simplify the denominator.

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Step 7.2.3.1

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

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

Step 7.2.3.2

Combine the numerators over the common denominator.

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

Step 7.2.3.3

Subtract $1$ from $2$ .

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

Step 7.2.3.4

Apply the product rule to $\frac{1}{2}$ .

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

Step 7.2.3.5

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

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Step 7.2.3.5.1

Multiply $- 1$ by $- 1$ .

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

Step 7.2.3.5.2

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

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

Step 7.2.3.6

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

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

Step 7.2.3.7

Combine the numerators over the common denominator.

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

Step 7.2.3.8

Add $2$ and $1$ .

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

Step 7.2.3.9

Apply the product rule to $\frac{3}{2}$ .

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

Step 7.2.3.10

One to any power is one.

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

Step 7.2.3.11

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

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

Step 7.2.3.12

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

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

Step 7.2.3.13

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

$f' (- \frac{1}{2}) = \frac{\frac{- 4}{2}}{\frac{1}{4} \cdot \frac{9}{4}}$

Step 7.2.4

Combine fractions.

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Step 7.2.4.1

Divide $- 4$ by $2$ .

$f' (- \frac{1}{2}) = \frac{- 2}{\frac{1}{4} \cdot \frac{9}{4}}$

Step 7.2.4.2

Multiply $\frac{1}{4}$ by $\frac{9}{4}$ .

$f' (- \frac{1}{2}) = \frac{- 2}{\frac{9}{4 \cdot 4}}$

Step 7.2.4.3

Multiply $4$ by $4$ .

$f' (- \frac{1}{2}) = \frac{- 2}{\frac{9}{16}}$

Step 7.2.5

Multiply the numerator by the reciprocal of the denominator.

$f' (- \frac{1}{2}) = - 2 (\frac{16}{9})$

Step 7.2.6

Multiply $- 2 (\frac{16}{9})$ .

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Step 7.2.6.1

Combine $- 2$ and $\frac{16}{9}$ .

$f' (- \frac{1}{2}) = \frac{- 2 \cdot 16}{9}$

Step 7.2.6.2

Multiply $- 2$ by $16$ .

$f' (- \frac{1}{2}) = \frac{- 32}{9}$

Step 7.2.7

Move the negative in front of the fraction.

$f' (- \frac{1}{2}) = - \frac{32}{9}$

Step 7.2.8

The final answer is $- \frac{32}{9}$ .

$- \frac{32}{9}$

Step 7.3

At $x = - \frac{1}{2}$ the derivative is $- \frac{32}{9}$ . Since this is negative, the function is decreasing on $(- 1, 0)$ .

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

Step 8

Substitute a value from the interval $(0, 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

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

Step 8.2

Simplify the result.

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Step 8.2.1

Combine fractions.

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Step 8.2.1.1

Remove parentheses.

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

Step 8.2.1.2

Combine $4$ and $\frac{1}{2}$ .

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

Step 8.2.2

Simplify the denominator.

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Step 8.2.2.1

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

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

Step 8.2.2.2

Combine the numerators over the common denominator.

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

Step 8.2.2.3

Add $2$ and $1$ .

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

Step 8.2.2.4

Apply the product rule to $\frac{3}{2}$ .

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

Step 8.2.2.5

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

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

Step 8.2.2.6

Combine the numerators over the common denominator.

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

Step 8.2.2.7

Subtract $1$ from $2$ .

$f' (\frac{1}{2}) = \frac{\frac{4}{2}}{\frac{3^{2}}{2^{2}} \cdot {(\frac{1}{2})}^{2}}$

Step 8.2.2.8

Apply the product rule to $\frac{1}{2}$ .

$f' (\frac{1}{2}) = \frac{\frac{4}{2}}{\frac{3^{2}}{2^{2}} \cdot \frac{1^{2}}{2^{2}}}$

Step 8.2.2.9

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

$f' (\frac{1}{2}) = \frac{\frac{4}{2}}{\frac{9}{2^{2}} \cdot \frac{1^{2}}{2^{2}}}$

Step 8.2.2.10

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

$f' (\frac{1}{2}) = \frac{\frac{4}{2}}{\frac{9}{4} \cdot \frac{1^{2}}{2^{2}}}$

Step 8.2.2.11

One to any power is one.

$f' (\frac{1}{2}) = \frac{\frac{4}{2}}{\frac{9}{4} \cdot \frac{1}{2^{2}}}$

Step 8.2.2.12

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

$f' (\frac{1}{2}) = \frac{\frac{4}{2}}{\frac{9}{4} \cdot \frac{1}{4}}$

Step 8.2.3

Combine fractions.

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Step 8.2.3.1

Divide $4$ by $2$ .

$f' (\frac{1}{2}) = \frac{2}{\frac{9}{4} \cdot \frac{1}{4}}$

Step 8.2.3.2

Multiply $\frac{9}{4}$ by $\frac{1}{4}$ .

$f' (\frac{1}{2}) = \frac{2}{\frac{9}{4 \cdot 4}}$

Step 8.2.3.3

Multiply $4$ by $4$ .

$f' (\frac{1}{2}) = \frac{2}{\frac{9}{16}}$

Step 8.2.4

Multiply the numerator by the reciprocal of the denominator.

$f' (\frac{1}{2}) = 2 (\frac{16}{9})$

Step 8.2.5

Multiply $2 (\frac{16}{9})$ .

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Step 8.2.5.1

Combine $2$ and $\frac{16}{9}$ .

$f' (\frac{1}{2}) = \frac{2 \cdot 16}{9}$

Step 8.2.5.2

Multiply $2$ by $16$ .

$f' (\frac{1}{2}) = \frac{32}{9}$

Step 8.2.6

The final answer is $\frac{32}{9}$ .

$\frac{32}{9}$

Step 8.3

At $x = \frac{1}{2}$ the derivative is $\frac{32}{9}$ . Since this is positive, the function is increasing on $(0, 1)$ .

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

Step 9

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 9.1

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

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

Step 9.2

Simplify the result.

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Step 9.2.1

Simplify the expression.

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Step 9.2.1.1

Remove parentheses.

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

Step 9.2.1.2

Multiply $4$ by $2$ .

$f' (2) = \frac{8}{{(1 + 2)}^{2} {(1 - (2))}^{2}}$

Step 9.2.2

Simplify the denominator.

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Step 9.2.2.1

Add $1$ and $2$ .

$f' (2) = \frac{8}{3^{2} {(1 - (2))}^{2}}$

Step 9.2.2.2

Multiply $- 1$ by $2$ .

$f' (2) = \frac{8}{3^{2} {(1 - 2)}^{2}}$

Step 9.2.2.3

Subtract $2$ from $1$ .

$f' (2) = \frac{8}{3^{2} {(- 1)}^{2}}$

Step 9.2.2.4

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

$f' (2) = \frac{8}{9 {(- 1)}^{2}}$

Step 9.2.2.5

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

$f' (2) = \frac{8}{9 \cdot 1}$

Step 9.2.3

Multiply $9$ by $1$ .

$f' (2) = \frac{8}{9}$

Step 9.2.4

The final answer is $\frac{8}{9}$ .

$\frac{8}{9}$

Step 9.3

At $x = 2$ the derivative is $\frac{8}{9}$ . Since this is positive, the function is increasing on $(1, \infty)$ .

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

Step 10

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

Increasing on: $(0, 1), (1, \infty)$

Decreasing on: $(- \infty, - 1), (- 1, 0)$

Step 11