Calculus Examples

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

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

[- 6, 6]

$[- 6, 6]$

Step 1

Find the derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Rewrite

\frac{1}{x}

$\frac{1}{x}$ as

x^{- 1}

$x^{- 1}$ .

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

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

Step 1.1.2

Differentiate using the Power Rule which states that

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

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

n x^{n - 1}

$n x^{n - 1}$ where

n = - 1

$n = - 1$ .

- x^{- 2}

$- x^{- 2}$

Step 1.1.3

Rewrite the expression using the negative exponent rule

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

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

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

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

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

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

Step 1.2

The first derivative of

f (x)

$f (x)$ with respect to

x

$x$ is

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

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

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

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

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

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

Step 2

Find if the derivative is continuous on

[- 6, 6]

$[- 6, 6]$ .

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Step 2.1

To find whether the function is continuous on

[- 6, 6]

$[- 6, 6]$ or not, find the domain of

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

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

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Step 2.1.1

Set the denominator in

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

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

0

$0$ to find where the expression is undefined.

x^{2} = 0

$x^{2} = 0$

Step 2.1.2

Solve for

x

$x$ .

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Step 2.1.2.1

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

x = \pm \sqrt{0}

$x = \pm \sqrt{0}$

Step 2.1.2.2

Simplify

\pm \sqrt{0}

$\pm \sqrt{0}$ .

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Step 2.1.2.2.1

Rewrite

0

$0$ as

0^{2}

$0^{2}$ .

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

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

Step 2.1.2.2.2

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

x = \pm 0

$x = \pm 0$

Step 2.1.2.2.3

Plus or minus

0

$0$ is

0

$0$ .

x = 0

$x = 0$

x = 0

$x = 0$

x = 0

$x = 0$

Step 2.1.3

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

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

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

Set-Builder Notation:

{x | x \neq 0}

${x | x \neq 0}$

Interval Notation:

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

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

Set-Builder Notation:

{x | x \neq 0}

${x | x \neq 0}$

Step 2.2

f' (x)

$f' (x)$ is not continuous on

[- 6, 6]

$[- 6, 6]$ because

0

$0$ is not in the domain of

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

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

The function is not continuous.

Step 3

The function is not differentiable on

[- 6, 6]

$[- 6, 6]$ because the derivative

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

$- \frac{1}{x^{2}}$ is not continuous on

[- 6, 6]

$[- 6, 6]$ .

The function is not differentiable.

Step 4

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