Question

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

Answer 1

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(x) = 4^x-8`

Differentiate:

`f'(x) = (d(4^x))/dx - (d(8))/dx`

The derivative of a constant is 0:

`f'(x) = (d(4^x))/dx`

Let `y = 4^x`

Use logarithmic differentiation:

`(d(ln(y)))/dx = (d(ln(4^x)))/dx = ln(4)dx/dx`

`1/ydy/dx = ln(4)`

`dy/dx = ln(4)y`

`dy/dx = ln(4)4^x`

Set equal to 0:

`ln(4)4^x=0 larr`This cannot happen.

Therefore, there is no minimum or maximum.

Please look at the graph:

graph{4^x-8 [-22.8, 22.81, -11.4, 11.42]}

Please observe that it has two asymptotes but no local extrema. The graph tends to toward -8 but that is not a minimum.