Precalculus Examples

$p (t) = 500 \cdot 2 t$

Step 1

Consider the difference quotient formula.

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

Step 2

Find the components of the definition.

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Step 2.1

Evaluate the function at $x = t + h$ .

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Step 2.1.1

Replace the variable $t$ with $t + h$ in the expression.

$p (t + h) = 500 \cdot (2 (t + h))$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Apply the distributive property.

$p (t + h) = 500 \cdot (2 t + 2 h)$

Step 2.1.2.2

Apply the distributive property.

$p (t + h) = 500 (2 t) + 500 (2 h)$

Step 2.1.2.3

Multiply.

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Step 2.1.2.3.1

Multiply $2$ by $500$ .

$p (t + h) = 1000 t + 500 (2 h)$

Step 2.1.2.3.2

Multiply $2$ by $500$ .

$p (t + h) = 1000 t + 1000 h$

$p (t + h) = 1000 t + 1000 h$

Step 2.1.2.4

The final answer is $1000 t + 1000 h$ .

$1000 t + 1000 h$

$1000 t + 1000 h$

$1000 t + 1000 h$

Step 2.2

Find the components of the definition.

$p (t + h) = 1000 t + 1000 h$

$p (t) = 1000 t$

$p (t + h) = 1000 t + 1000 h$

$p (t) = 1000 t$

Step 3

Plug in the components.

$\frac{p (t + h) - p t}{h} = \frac{1000 t + 1000 h - (1000 t)}{h}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $- 1$ by $1000$ .

$\frac{1000 t + 1000 h - 1000 t}{h}$

Step 4.1.2

Subtract $1000 t$ from $1000 t$ .

$\frac{1000 h + 0}{h}$

Step 4.1.3

Add $1000 h$ and $0$ .

$\frac{1000 h}{h}$

$\frac{1000 h}{h}$

Step 4.2

Cancel the common factor of $h$ .

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Step 4.2.1

Cancel the common factor.

$\frac{1000 h}{h}$

Step 4.2.2

Divide $1000$ by $1$ .

$1000$

$1000$

$1000$

Step 5

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