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

Feb 24, 2018

This function has no local extrema.

#### Explanation:

At a local extremum, we must have $f p r i m e \left(x\right) = 0$
Now,
$f p r i m e \left(x\right) = \left({x}^{2} + 8 x + 3\right) {e}^{x} + 8$

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

Since $g p r i m e \left(x\right) = \left({x}^{2} + 10 x + 11\right) {e}^{x}$, the extrema of $g \left(x\right)$ are at the points where ${x}^{2} + 10 x + 11 = 0$, i.e. at $x = - 5 \pm \sqrt{14}$. Since $g \left(x\right) \to \infty$ 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 $g \left(- 5 + \sqrt{14}\right) \approx - 1.56$, so that the minimum value of $f p r i m e \left(x\right) \approx 6.44$ - so that it can never reach zero.