Applications of Differentiation

Question

Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a single category list the points in increasing order in x value and enter N in any blank that you don't need to use.

f(x) = 8e^(-x) - 8e^(-2x) , [0,1]

Absolute maxima

x =             y = 

x =             y = 

x =             y = 

Absolute minima

x =         y = 

x =         y = 

x =        y = 

Answer 1

The standard way you will see this solved is to take the derivative and set it equal to 0, so I will give you an alternative solution.

f(x) = 8e^(-x) - 8e-2x = -8e^(-2x) + 8e^(-2x) - 2 + 2 =

-8(e^(-x) - 1/2)^2 + 2

This is maximized at e^(-x) = 1/2, or e^x = 2, or x = ln(2), and has a maximum value of 2.

Note that ln(2) is in[0, 1]

e^-x > e^(-2x) on (0, ∞), so f(x) > 0 on (0, ∞)

f(0) = 0. This is in [0, 1]

Thus, the absolute maximum is

x = ln(2) y = 2

The absolute minimum is

x = 0 y = 0

Answer 2

f'(x) = -8e^-x + 16 e^-2x = - 8e^-x(1 - 2e^-x) = 0; - 8e^-x < 0 for all x; so, 1 - 2e^-x = 0; e^-x = 1/2; x = ln2; 0 < ln2 < 1;

f(0) = 8 - 8 = 0; abs min

f(ln2) = 8e^-ln2 - 8e^-2ln2 = 4 - 2 = 2 abs max

f(1) = 8e^-1 - 8e^-2 = 1.8604