Let $f(x)=\int_0^{x^2}\frac{t^2-8t+15}{e^t}\,dt$, $x\in\mathbf R$. Then the numbers of local maximum and local minimum points of $f$, respectively, are :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $f'(x)=2x\,(x^2-3)(x^2-5)\,e^{-x^2}$. The sign of $f'$ from left to right is $-,+,-,+,-,+$ across $x=-\sqrt5,-\sqrt3,0,\sqrt3,\sqrt5$. So $-\sqrt5,0,\sqrt5$ are local minima and $\pm\sqrt3$ are local maxima: 2 maxima and 3 minima. Answer (1).

Q. Let $f(x)=\int_0^{x^2}\frac{t^2-8t+15}{e^t}\,dt$, $x\in\mathbf R$. Then the numbers of local maximum and local minimum points of $f$, respectively, are :

Explanation

Reviewed by Dr. Abubacker

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