No. of solutions of $f(x)=f'(x)$?

Question

Let $f:[0,1] \to \Bbb R$ be a fixed continuous function such that $f$ is differentiable on $(0,1)$ and $f(0)=f(1)$. Then the equation $f(x)=f'(x)$ admits

1. No solution $x \in (0,1)$
2. More than one solution $x \in (0,1)$
3. Exactly one solution $x \in (0,1)$
4. At least one solution $x \in (0,1)$

As I have tried taking $f(x)=0$ on $[0,1]$ ruled out options 1 and 2 and by Rolle's Theorem there exists $c\in (0,1)$ such that $f'(c)=0$. Then I thought to construct function $g(x)=f(x)-f'(x)$ to check zeros but I'm stuck because $f'(x)$ need to be continuous.

Can anyone give some hint to proceed further?

Answer 1

I don't think you can guarantee any of the choices. Below are three possibilities for $f(x)$ which have

1. $0$ solutions, 2. $1$ solution, 3. $\infty$ solutions.

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1. $f(x)=4$
2. $f(x)=(x-\frac12)^2$
3. $f(x)=0$

Answer 2

1. No solution $x\in(0,1)$. Yes, this can happen. Consider the constant polynomial $p(x)=5$. We have $p(0)=p(1)=5$ and $p(x)\neq p^\prime(x)$ for all $x\in(0,1)$.

2. More than one solution $x\in(0,1)$. Yes, this can happen. Build a polynomial function $p(x)$, with degree $8$, whose roots are $r_1=0$, $r_2=r_3=0.25$, $r_4=r_5=0.50$, $r_6=r_7=0.75$ and $r_8=1$. Then $p(r_2)=p^\prime(r_2)=0$, $p(r_4)=p^\prime(r_4)=0$ and $p(r_6)=p^\prime(r_6)=0$. Use Vieta's formulas.

3. Exactly one solution $x\in(0,1)$. Consider $p(x)=x^2-x$. We have $p(0)=p(1)=0$ and $\frac{3-\sqrt{5}}{2}$ root of $p(x)=p^\prime(x)$. As pointed out by lisyarus.

4. At least one solution $x\in(0,1)$ . From the first example given above, this statement is not true.