Question

A theorem states that if f is continuous on an interval I that contains one local extremum at c and if a local minimum occurs at c, then f(c) is the absolute minimum of f on I, and if a local maximum occurs at c, then f(c) is the absolute maximum of f on I. Verify that the following function satisfies the conditions of this theorem on its domain. Then, find the location and value of the absolute extremum guaranteed by the theorem.

f(x)= -xe^-x/4 The function f(x) = -xe^-x/4 has an absolute extremum of Q at x=.?

Answer 1

Explanation:

To find the extremum of the function, you need to take the first derivative.
`f'(x) = (-x)'e^(-x/4) + (-x) (e^(-x/4))'`

`= e^(-x/4)((x-4)/(4) )`

This derivative = 0 if and only if x - 4 = 0, hence the extremum is at x = 4

To consider if it is local max or min, you need to consider the act of the function before and after x = 4 by making a table.

`-\infty` 4
`+\infty`

f(x) - 0 +

`\searrow`
`\\earrow`

Hence x =4 is a local min.