# How do you determine the minimum value of the function f(x)= 4^x -8?

Mar 30, 2018

Compute the first derivative
Set that equal to zero and the value(s) of x.
Use the second derivative test to determine whether it is a minimum.

#### Explanation:

Given $f \left(x\right) = {4}^{x} - 8$

Differentiate:

$f ' \left(x\right) = \frac{d \left({4}^{x}\right)}{\mathrm{dx}} - \frac{d \left(8\right)}{\mathrm{dx}}$

The derivative of a constant is 0:

$f ' \left(x\right) = \frac{d \left({4}^{x}\right)}{\mathrm{dx}}$

Let $y = {4}^{x}$

Use logarithmic differentiation:

$\frac{d \left(\ln \left(y\right)\right)}{\mathrm{dx}} = \frac{d \left(\ln \left({4}^{x}\right)\right)}{\mathrm{dx}} = \ln \left(4\right) \frac{\mathrm{dx}}{\mathrm{dx}}$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(4\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(4\right) y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(4\right) {4}^{x}$

Set equal to 0:

$\ln \left(4\right) {4}^{x} = 0 \leftarrow$This cannot happen.

Therefore, there is no minimum or maximum.