Question

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

Answer 1

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.