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

Answer:

Step-by-step explanation:

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

$= e^{-x/4}(\frac{x-4}{4} )$

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                +

                     $\searrow$                                         $\nearrow$

Hence x =4 is a  local min.