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

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Aicharya · Accepted Answer

$\begin{aligned}
& f(x)=\int_0^{x^2} \frac{t^2-8 t+15}{e^t} d t, x \in R \\ & f^{\prime}(x)=\frac{x^4-8 x^2+15}{e^{x^2}}(2 x)=0 \\ & \Rightarrow \quad \frac{2 \times\left(x^2-5\right)\left(x^2-3\right)}{e^{x^2}}=0 \\ & \Rightarrow x(x+\sqrt{5})(x-\sqrt{5})(x+\sqrt{3})(x-\sqrt{3})=0
\end{aligned}$
By using wavy curve method

Number of local maximum $=2$
Number of local minimum $=3$

Let $f(x)=\int_0^{x^2} \frac{\mathrm{t}^2-8 \mathrm{t}+15}{\mathrm{e}^{\mathrm{t}}} \mathrm{dt}, x \in…

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