Assume $f:[a,b] \to \mathbb R$ is continuous on the closed interval $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)=0$. Prove that for every $c \in \mathbb R$ there exists $x_0 \in (a,b)$ such that $f'(x_0)= cf(x_0)$.
Here is my work so far: Assume the case is true for $c = 1$. Then for any other $c$ let $g(x):=f(cx)$. Then since the case holds for $c=1$ there exists $x_0 \in [a/c, b/c]$ such that $f'(cx_0)=1/c f(cx_0)$. My inteuctor said that, while my effort for reducion does not exactly work (since we have to check that our motified value lies ina proper interval and check other things), I can make some similar argument to reduce the case to $c=1$. Some people suggested that I take a look at the logarithmic and exponential functions, but I am not sure how that would lead me anywhere.
Standard trick: consider $g(x) = e^{-cx} f(x)$. Then
$$g'(x) = e^{-cx} f'(x) - c e^{-cx} f(x) = e^{-cx} (f'(x) - cf(x)).$$
So you are asking if $g'(x)$ is zero at some point. But $g(a) = g(b) = 0$.