Intuitive understanding for $f(x)=f'(x)$ in $(a,b)$ when $f(a)=f(b)=0$

Question

Question:

Let $f(x)$ be a differentiable function and $f(a)=f(b)=0 \;(a\lt b)$ then in the interval $(a,b)$

1. $f(x)+f'(x)=0$ has at least one root.

2. $f(x)-f'(x)=0$ has at least one root.

3. $f(x)\times f'(x)=0$ has at least one root.

4. none of these

My Attempt:

Perhaps Method $1$ is the actual way to do it. But I want to understand it in an intuitive way i.e. by Method $2$.

Method $1$:

By Rolle's theorem, $f'(x)=0$ at least once in $(a,b)$, thus 3) is correct.

Let $g(x)=e^xf(x)$. By Rolle's theorem, 1) is correct.

Let $h(x)=e^{-x}f(x)$. By Rolle's theorem, 2) is correct.

Method $2$:

Since $f(a)=f(b)=0$, and $f(x)$ is differentiable, (I am assuming $f(x)$ to be concave downwards in $(a,b)$), thus $f(x)$ is increasing from $a$, reaching a peak and coming down at $b$.

That means $f'(x)$ is varying from $-\infty$ at $a$ to zero at the peak to $+\infty$ at $b$.

Thus, $f'(x)$ here has covered all real values in $(a,b)$. For $p\in(a,b)$, let $f'(p)=q$

For 2) to be correct, why are we saying that even $f(p)=q$?

Why can't $f(x)$ have some other value at $p$?

Answer 1

The Question asks about the Intuition, hence I will give Pictorial Intuitive Heuristics about this.

When $f(x)$ has Multiple Crossings (with Multiple Maxima & Minima Points) about the x-axis, we can always get a sub-interval $(A,B) \subset (a,b)$ where there is only 1 Maxima. In case there is only 1 Minima, we can consider Negative ( that is $-f(x)$ ) & get the Maxima.

Hence, without loss of generality, we can take that $f(x)$ has 1 Maxima between $A$ & $B$.
In this case, the Derivative is Positive $(P)$ at $A$ & Negative $(-N)$ at $B$. Shown in the Picture is the function $f(x)$ starting at $(A,0)$ & ending at $(B,0)$ & the Derivative starting at $(A,P)$ & ending at $(B,-N)$.

OPTION 3 : Derivative starts Positive $P$ & ends Negative $-N$ , hence there is a root of the Derivative in between (Center Circle).

OPTION 1 : The Sum will start at $0+P$ & then go to $0+(-N)$. It has to go to 0 somewhere in between. Hence there is at least 1 root of the Sum (right Circle).

Alternatively, This occurs where The Derivative has Exact Negative value of $f(x)$ to cancel the Sum too (right Circle).

OPTION 2 : Of course, there will a Point where the function and the Derivative will be equal, that will be the root of the Difference (left Circle).

Alternatively, The Difference will start at Negative value $(0-P)$ & then go to Positive value $(0-(-N))$. It has to go to 0 somewhere in between. Hence there is at least 1 root of the Difference too (left Circle).