Find m so that f has one local extreme point. Let $f\colon (-1,1) \to \mathbb R $ and $ f(x)=e^x(x^2+x+m) $.
The function f has only one extreme local point if and only if m belongs to the set:
a) (-5, 1);     b) {-5, 1};     c) [-5, 1);    d) ${\frac{5}{4}}$.
I tried to calcule the derivate and then $\Delta = 0$. I got $m = \frac{5}{4}$, but I'm not sure I set the right condition.
$f'(x)=e^x(x^2+3x+m+1)$
$\Delta=5-4m=0 $
$ m=\frac{5}{4}$
 A: This problem contains a little "trap" (which I suspect the poser enjoyed placing there).  The function has a domain restriction which is an open interval:  this will mean that absolute extrema are not guaranteed, since the Extreme Value Theorem applies to closed intervals.  One also needs to be wary that the expression for the function may have relative extrema which are not in the domain $ \ (-1 \ , \ 1) \ $ of the function.
So saying, the derivative function has its zeroes (when they exist) at $ \ x \ = \  \ -\frac32 \ \pm \ \frac{\sqrt{5-4m}}{2} \ \ . $  When the discriminant is zero, the single zero is not in the stated domain, so $ \ m \ = \ \frac54 \ $ will not be a correct choice. [In fact, the derivative function is then $ \ f'(x) \ = \ e^x· \left(x + \frac32  \right)^2 \ \ , $ which means that $ \ x \ = \ -\frac32 \ $ is an inflection point for the function curve.]  Setting   $ \ x \ = \ -\frac32 \ + \ \frac{\sqrt{5-4m}}{2} \ \  $ equal to $ \ +1 \ $ or $ \ -1 \ $ will give us bounds of the value of $ \ m \ \ , $ which of course will not be included in the solution interval.  We also see from $ \ x \ = \ -\frac32 \ - \ \frac{\sqrt{5-4m}}{2} \ \  $ that the second extremum (a local maximum) of the function expression remains outside of $ \ (-1 \ , \ 1)  \ \ . $
