1. Rewrite the given expression

We need the minimum number of solutions of $f(x)g''(x)+f'(x)g'(x)=0.$

Notice that $\frac{d}{dx}\big(f(x)g'(x)\big)=f'(x)g'(x)+f(x)g''(x).$ So the equation becomes $\big(f(x)g'(x)\big)'=0.$

Thus, we need the minimum number of zeros of the derivative of $H(x)=f(x)g'(x).$ By Rolle's theorem, if we can find intervals where $H$ takes equal values at the endpoints, then $H'(x)=0$ somewhere inside.

2. Use the fact that $f$ and $g$ are even

Since $f$ and $g$ are even:

• $f(-x)=f(x)$,
• $g(-x)=g(x)$.

Hence:

• $f'$ and $g'$ are odd,
• so $g'(0)=0$.

Also given: $f\left(\frac14\right)=0,\quad f\left(\frac12\right)=0,\quad f(1)=1,$ so by evenness, $f\left(-\frac14\right)=0,\quad f\left(-\frac12\right)=0.$

And given: $g\left(\frac34\right)=0,\quad g(1)=2,$ so by evenness, $g\left(-\frac34\right)=0,\quad g(-1)=2.$

3. First find zeros of $g'$ using Rolle's theorem

Apply Rolle's theorem to $g$ on intervals where endpoint values are equal.

• On $\left[-1,-\frac34\right]$: $g(-1)=2,\quad g\left(-\frac34\right)=0$ not equal, so no direct Rolle.

• On $\left[-\frac34,\frac34\right]$: $g\left(-\frac34\right)=0=g\left(\frac34\right),$ so there exists $c_1\in\left(-\frac34,\frac34\right)$ such that $g'(c_1)=0.$

Also, since $g$ is even, $g'(0)=0$ already.

Now use Mean Value / Rolle more carefully on subintervals created by the values of $g$:

• On $\left[\frac34,1\right]$, $g\left(\frac34\right)=0$, $g(1)=2$; by MVT there exists $a\in\left(\frac34,1\right)$ with $g'(a)>0$.
• By oddness, on $\left(-1,-\frac34\right)$ there exists $-a$ with $g'(-a)<0$.
• Since $g'(0)=0$ and $g'$ is continuous, this alone does not force extra zeros beyond $0$.

But for our equation we need zeros of $H=f g'$, and these can come from either $f=0$ or $g'=0$.

4. Zeros of $H(x)=f(x)g'(x)$

We know:

• $f\left(-\frac12\right)=0$,
• $f\left(-\frac14\right)=0$,
• $f\left(\frac14\right)=0$,
• $f\left(\frac12\right)=0$.

Therefore, $H\left(-\frac12\right)=H\left(-\frac14\right)=H\left(\frac14\right)=H\left(\frac12\right)=0.$

Also, because $g$ is even, $g'(0)=0$, so $H(0)=f(0)g'(0)=0.$

Thus $H$ has at least the five zeros $-\frac12,\ -\frac14,\ 0,\ \frac14,\ \frac12.$

5. Apply Rolle's theorem to $H$

Since $H$ is differentiable, between every two consecutive zeros of $H$, there is at least one point where $H'(x)=0$.

The five distinct zeros above create four intervals: $\left(-\frac12,-\frac14\right),\quad \left(-\frac14,0\right),\quad \left(0,\frac14\right),\quad \left(\frac14,\frac12\right).$

Hence, by Rolle's theorem, there exist at least four distinct points in these intervals where $H'(x)=0.$ That is, $f(x)g''(x)+f'(x)g'(x)=0$ has at least 4 solutions in $(-2,2)$.

6. Why this minimum can actually be 4

The information given guarantees these 5 zeros of $H$, and hence 4 zeros of $H'$. No additional zeros are forced by the data.

So the minimum number of solutions is $4.$

7. Comparison with stored answer

Derived answer: $4$.

Stored correct answer: $4$.

They agree.