Let \(f(x)=\int_0^{x^2} \frac{t^2-8 t+15}{e^t} d t, x \in \mathbf{R}\). Then the numbers of local maximum and local mini…

\(f\left(x\right)={\int }_{0}^{{x}^{2}}\frac{{t}^{2}-8t+15}{{e}^{t}}dt,x\in R\\ {f}^{'}\left(x\right)=\left(\frac{{x}^{4}-8{x}^{2}+15}{{e}^{{x}^{2}}}\right)\left(2x\right)\\ =\frac{\left({x}^{2}-3\right)\left({x}^{2}-5\right)(2x)}{{\mathrm{e}}^{{\mathrm{x}}^{2}}}\\ =\frac{(x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})2x}{{e}^{{x}^{2}}}\)

using first derivative approach :-

\(\text{Maxima at }x\in {-\sqrt{3},\sqrt{3}}\\ \text{Minima at }\mathrm{x}\in {-\sqrt{5},0,\sqrt{5}}\)

number of local maxima points=\(2\)

number of local minima points=\(3\)