What are the local extrema, if any, of #f(x)= (x^2 + 6x-3)*e^x + 8x –8#?

1 Answer
Feb 24, 2018

Answer:

This function has no local extrema.

Explanation:

At a local extremum, we must have #f prime(x)=0#
Now,
#f prime (x) = (x^2+8x+3)e^x+8#

Let us consider whether this can vanish. For this to happen, the value of #g(x) = (x^2+8x+3)e^x # must be equal to -8.

Since #g prime(x) = (x^2+10x+11)e^x#, the extrema of #g(x)# are at the points where #x^2+10x+11=0#, i.e. at #x=-5 pm sqrt{14}#. Since #g(x) to infty# and 0 as #x to pm infty# respectively, it is easy to see that the minimum value will be at #x = -5+sqrt{14}#.

We have #g(-5+sqrt{14}) ~~ -1.56#, so that the minimum value of #f prime (x) ~~ 6.44# - so that it can never reach zero.