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

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Answer 1 · 2018-03-30T17:05:21

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.

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.