Finding Relative Max

Question

If the derivative of f is given by f'(x)=e^x-3x^2 at which of the following values of x does f have a relative maximum value?

a) -0.46

b) 0.20

c) 0.91

d) 0.95

e) 3.73

Answer 1

To find a local minimum or maximum, find the critical points by setting f ' = zero.

e^x - 3x² = 0

e^x = 3x²

The easiest way to solve this is to graph it and look for the x-intercepts.

x = -0.458962

x = 0.9100076

x = 3.733079

Putting these points on a number line we can perform the first derivative test by evaluating points in each increment they divide the line into:

The red circles represent the value of the derivative for points in those intervals

f '(-1) = -2.6 (a negative - meaning the function is decreasing on this interval)

f '(0) = 1 (a positive - meaning the function is increasing on this interval)

f '(1) = -0.3 (a negative - meaning the function is decreasing on this interval)

f '(4) = 6.6 (a positive - meaning the function is increasing on this interval)

So x = -0.458962 must be a local minimum

And x = 3.733079 must be a local minimum

That leaves the local maximum at x = 0.9100076

To find that maximum value we would need to know what the function is and we only know the derivative, The function is the antiderivative of f '(x) or f(x) = ex -x3 + C (where C is some constant). Since we don't have any data point that the function goes through, we cannot determine the function or the local maximum vale. But it appears by looking at the answers that the x-value of the local max is what is being asked for.