Precalculus Examples

$K_{c} = 4 \sqrt{4 v + 1}$

Step 1

Write $K_{c} = 4 \sqrt{4 v + 1}$ as a function.

$f (v) = 4 \sqrt{4 v + 1}$

Step 2

Consider the difference quotient formula.

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

Step 3

Find the components of the definition.

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Step 3.1

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

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Step 3.1.1

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

$f (v + h) = 4 \sqrt{4 (v + h) + 1}$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Apply the distributive property.

$f (v + h) = 4 \sqrt{4 v + 4 h + 1}$

Step 3.1.2.2

The final answer is $4 \sqrt{4 v + 4 h + 1}$ .

$4 \sqrt{4 v + 4 h + 1}$

Step 3.2

Find the components of the definition.

$f (v + h) = 4 \sqrt{4 v + 4 h + 1}$

$f (v) = 4 \sqrt{4 v + 1}$

$f (v + h) = 4 \sqrt{4 v + 4 h + 1}$

$f (v) = 4 \sqrt{4 v + 1}$

Step 4

Plug in the components.

$\frac{f (v + h) - f v}{h} = \frac{4 \sqrt{4 v + 4 h + 1} - (4 \sqrt{4 v + 1})}{h}$

Step 5

Simplify the numerator.

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Step 5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 v + 4 h + 1}$ as ${(4 v + 4 h + 1)}^{\frac{1}{2}}$ .

$\frac{4 {(4 v + 4 h + 1)}^{\frac{1}{2}} - 1 \cdot 4 \sqrt{4 v + 1}}{h}$

Step 5.2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 v + 1}$ as ${(4 v + 1)}^{\frac{1}{2}}$ .

$\frac{4 {(4 v + 4 h + 1)}^{\frac{1}{2}} - 1 \cdot 4 {(4 v + 1)}^{\frac{1}{2}}}{h}$

Step 5.3

Multiply $- 1$ by $4$ .

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

Step 5.4

Factor $4$ out of $4 {(4 v + 4 h + 1)}^{\frac{1}{2}} - 4 {(4 v + 1)}^{\frac{1}{2}}$ .

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Step 5.4.1

Factor $4$ out of $4 {(4 v + 4 h + 1)}^{\frac{1}{2}}$ .

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

Step 5.4.2

Factor $4$ out of $- 4 {(4 v + 1)}^{\frac{1}{2}}$ .

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

Step 5.4.3

Factor $4$ out of $4 {(4 v + 4 h + 1)}^{\frac{1}{2}} + 4 (- {(4 v + 1)}^{\frac{1}{2}})$ .

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

Step 6