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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}$

claudiubagiu
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    I think your solution is correct. – 温泽海 May 04 '22 at 23:20
  • yep, you know that $e^x=0$ has no roots and so you want the quadratic to have exactly 1 root, which means the discriminant is zero ($\Delta=0$), all good :) – Henry Lee May 04 '22 at 23:33
  • The question asks for values of $ \ m \ $ for which there is only one extremum within the interval $ \ (-1 \ , \ 1) \ $ ; the derivative may have a second zero outside of that interval. [Note: $ \ m \ $ need not take on every value in the suggested intervals. For $ \ m \ = \ \frac54 \ $ , that function has no extremal points; there is a point of inflection at $ \ x \ = \ -\frac32 \ . $ ] –  May 05 '22 at 01:13

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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) \ \ . $