Calculus Examples

f (x) = - \frac{1}{x}

$f (x) = - \frac{1}{x}$ ,

- 2 \leq x \leq - 1

$- 2 \leq x \leq - 1$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Differentiate using the Product Rule which states that

\frac{d}{d x} [f (x) g (x)]

$\frac{d}{d x} [f (x) g (x)]$ is

f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]

$f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where

f (x) = - 1

$f (x) = - 1$ and

g (x) = \frac{1}{x}

$g (x) = \frac{1}{x}$ .

- \frac{d}{d x} [\frac{1}{x}] + \frac{1}{x} \frac{d}{d x} [- 1]

$- \frac{d}{d x} [\frac{1}{x}] + \frac{1}{x} \frac{d}{d x} [- 1]$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

Rewrite

\frac{1}{x}

$\frac{1}{x}$ as

x^{- 1}

$x^{- 1}$ .

- \frac{d}{d x} [x^{- 1}] + \frac{1}{x} \frac{d}{d x} [- 1]

$- \frac{d}{d x} [x^{- 1}] + \frac{1}{x} \frac{d}{d x} [- 1]$

Step 1.1.1.2.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = - 1$ .

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

Step 1.1.1.2.3

Multiply.

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Step 1.1.1.2.3.1

Multiply

$- 1$ by

$- 1$ .

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

Step 1.1.1.2.3.2

Multiply

$x^{- 2}$ by

$1$ .

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

Step 1.1.1.2.4

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- 1$ with respect to

$x$ is

$0$ .

$x^{- 2} + \frac{1}{x} \cdot 0$

Step 1.1.1.2.5

Simplify the expression.

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Step 1.1.1.2.5.1

Multiply

$\frac{1}{x}$ by

$0$ .

$x^{- 2} + 0$

Step 1.1.1.2.5.2

Add

$x^{- 2}$ and

$0$ .

$x^{- 2}$

Step 1.1.1.3

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.1.2

The first derivative of

$f (x)$ with respect to

$x$ is

$\frac{1}{x^{2}}$ .

$\frac{1}{x^{2}}$

Step 1.2

Set the first derivative equal to

$0$ then solve the equation

$\frac{1}{x^{2}} = 0$ .

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Step 1.2.1

Set the first derivative equal to

$0$ .

$\frac{1}{x^{2}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$1 = 0$

Step 1.2.3

Since

$1 \neq 0$ , there are no solutions.

No solution

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

Set the denominator in

$\frac{1}{x^{2}}$ equal to

$0$ to find where the expression is undefined.

$x^{2} = 0$

Step 1.3.2

Solve for

$x$ .

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Step 1.3.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 1.3.2.2

Simplify

$\pm \sqrt{0}$ .

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Step 1.3.2.2.1

Rewrite

$0$ as

$0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 1.3.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 1.3.2.2.3

Plus or minus

$0$ is

$0$ .

$x = 0$

Step 1.4

Evaluate

$- \frac{1}{x}$ at each

$x$ value where the derivative is

$0$ or undefined.

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Step 1.4.1

Evaluate at

$x = 0$ .

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Step 1.4.1.1

Substitute

$0$ for

$x$ .

$- \frac{1}{0}$

Step 1.4.1.2

The expression contains a division by

$0$ . The expression is undefined.

Undefined

Step 1.5

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 2

Evaluate at the included endpoints.

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Step 2.1

Evaluate at

$x = - 2$ .

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Step 2.1.1

Substitute

$- 2$ for

$x$ .

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

Step 2.1.2

Simplify.

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Step 2.1.2.1

Move the negative in front of the fraction.

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

Step 2.1.2.2

Multiply

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

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Step 2.1.2.2.1

Multiply

$- 1$ by

$- 1$ .

$1 (\frac{1}{2})$

Step 2.1.2.2.2

Multiply

$\frac{1}{2}$ by

$1$ .

$\frac{1}{2}$

Step 2.2

Evaluate at

$x = - 1$ .

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Step 2.2.1

Substitute

$- 1$ for

$x$ .

$- \frac{1}{- 1}$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Divide

$1$ by

$- 1$ .

$- - 1$

Step 2.2.2.2

Multiply

$- 1$ by

$- 1$ .

$1$

Step 2.3

List all of the points.

$(- 2, \frac{1}{2}), (- 1, 1)$

Step 3

Compare the

$f (x)$ values found for each value of

$x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest

$f (x)$ value and the minimum will occur at the lowest

$f (x)$ value.

Absolute Maximum:

$(- 1, 1)$

Absolute Minimum:

$(- 2, \frac{1}{2})$

Step 4