1. Given function

We need the number of local maxima and local minima points of $f(x)$.

2. Differentiate using Fundamental Theorem of Calculus and chain rule

Let

Then

So,

Factor the polynomial:

Hence

Since $e^{x^2}>0$ for all $x$, the sign of $f'(x)$ depends only on

Critical points are:

3. Sign analysis of $f'(x)$

We check the intervals:

• $(-\infty,-\sqrt5)$
• $(-\sqrt5,-\sqrt3)$
• $(-\sqrt3,0)$
• $(0,\sqrt3)$
• $(\sqrt3,\sqrt5)$
• $(\sqrt5,\infty)$

Take signs of factors:

(i) $x< -\sqrt5$

• $x<0$ so $2x<0$
• $x^2-3>0$
• $x^2-5>0$

Thus $f'(x)<0$.

(ii) $-\sqrt5<x<-\sqrt3$

• $2x<0$
• $x^2-3>0$
• $x^2-5<0$

Thus $f'(x)>0$.

(iii) $-\sqrt3<x<0$

• $2x<0$
• $x^2-3<0$
• $x^2-5<0$

Thus $f'(x)<0$.

(iv) $0<x<\sqrt3$

• $2x>0$
• $x^2-3<0$
• $x^2-5<0$

Thus $f'(x)>0$.

(v) $\sqrt3<x<\sqrt5$

• $2x>0$
• $x^2-3>0$
• $x^2-5<0$

Thus $f'(x)<0$.

(vi) $x>\sqrt5$

• $2x>0$
• $x^2-3>0$
• $x^2-5>0$

Thus $f'(x)>0$.

So the sign chart is:

across the critical points

4. Determine local maxima and minima

A local minimum occurs when $f'(x)$ changes from negative to positive.

A local maximum occurs when $f'(x)$ changes from positive to negative.

Now check each critical point:

• At $x=-\sqrt5$: $f'$ changes $- \to +$
  $\Rightarrow$ local minimum

• At $x=-\sqrt3$: $f'$ changes $+ \to -$
  $\Rightarrow$ local maximum

• At $x=0$: $f'$ changes $- \to +$
  $\Rightarrow$ local minimum

• At $x=\sqrt3$: $f'$ changes $+ \to -$
  $\Rightarrow$ local maximum

• At $x=\sqrt5$: $f'$ changes $- \to +$
  $\Rightarrow$ local minimum

Therefore,

• Number of local maxima $=2$
• Number of local minima $=3$

5. Compare with options

This corresponds to:

6. Comparison with stored correct answer

Stored correct answer is C, which matches our result.