If m and n respectively are the number of local maximum and local...

Given $f\left(x\right) = {∫}_{0}^{{x}^{2}}\dfrac{{t}^{2} - 5t + 4}{2 + {e}^{t}}dt$

We know that,

$\dfrac{d}{dx}{∫}_{u\left(x\right)}^{v\left(x\right)}f\left(v\right)dx = f\left(v\left(x\right)\right)\dfrac{d}{dx}v\left(u\right) - f\left(u\left(x\right)\right)\dfrac{d}{dx}u\left(x\right)$

Using above formula we get ${f}^{'}\left(x\right) = \left(\dfrac{{x}^{4} - 5{x}^{2} + 4}{2 + {e}^{{x}^{2}}}\right) \cdot 2x$

For critical points ${f}^{'}\left(x\right) = 0$

So, $\dfrac{x\left({x}^{2} - 1\right)\left({x}^{2} - 4\right)}{2 + {e}^{{x}^{2}}} = 0$

$⇒ x\left(x + 1\right)\left(x - 1\right)\left(x + 2\right)\left(x - 2\right) = 0$

Now by first derivative test we get,

Here points of maxima are $\left\{- 1, 1\right\}$ & points of minima are \left\{- 2, 0, 2\right\}$