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Algebra Examples

f (x) = x

$f (x) = x$ ,

x = 0

$x = 0$

Step 1

Consider the difference quotient formula.

\frac{f (x + h) - f (x)}{h}

$\frac{f (x + h) - f (x)}{h}$

Step 2

Find the components of the definition.

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Step 2.1

Evaluate the function at

x = x + h

$x = x + h$ .

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Step 2.1.1

Replace the variable

x

$x$ with

x + h

$x + h$ in the expression.

f (x + h) = x + h

$f (x + h) = x + h$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Remove parentheses.

f (x + h) = x + h

$f (x + h) = x + h$

Step 2.1.2.2

The final answer is

x + h

$x + h$ .

x + h

$x + h$

x + h

$x + h$

x + h

$x + h$

Step 2.2

Find the components of the definition.

f (x + h) = x + h

$f (x + h) = x + h$

f (x) = x

$f (x) = x$

f (x + h) = x + h

$f (x + h) = x + h$

f (x) = x

$f (x) = x$

Step 3

Plug in the components.

\frac{f (x + h) - f (x)}{h} = \frac{x + h - (x)}{h}

$\frac{f (x + h) - f (x)}{h} = \frac{x + h - (x)}{h}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Subtract

x

$x$ from

x

$x$ .

\frac{h + 0}{h}

$\frac{h + 0}{h}$

Step 4.1.2

Add

h

$h$ and

0

$0$ .

\frac{h}{h}

$\frac{h}{h}$

\frac{h}{h}

$\frac{h}{h}$

Step 4.2

Cancel the common factor of

h

$h$ .

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Step 4.2.1

Cancel the common factor.

\frac{h}{h}

$\frac{h}{h}$

Step 4.2.2

Rewrite the expression.

1

$1$

1

$1$

1

$1$

Step 5

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[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$