Prove that $f'(c ) = \lambda f(c )$ Suppose $f$ is continuous on $[a,b]$ and $f$ is differentiable on $(a,b)$ with $f(a) = f(b) = 0$. Prove that for every real $\lambda, \exists c \in (a,b)$ s.t. $f'(c ) = \lambda f(c )$. 
Hint: Apply Rolle to $f(x)e^{g(x)}$ for suitable $g$. 
Now, for this problem, I coulnd't come up with anything as I exactly don't understand how/where to apply the hint. All I know from the Rolle's theorem is that $\exists c \in (a,b)$ s.t. $f'(c ) = 0$ after pretty much all the given conditions above. Any help would be much appreciated. 
 A: Rationale for $g(x)=-\lambda x$:
You need a lambda in the expression somewhere. $g=\lambda$ wouldn't work, clearly, so try the next simplest thing, a linear function $g=\lambda x$.
To continue a little bit, if $h(x)=f(x) e^{g(x)}$, Rolle's means that since $h(a) = f(a) e^{g(a)}= h(b) = f(b) e^{g(b)} = 0$, there is a $c$ such that $h'(c) = 0$.
You should be able to fill in the details.
Rationale for $f(x) e^{g(x)}$:
Why is there the hint in the first place?
This is a trickier one and mostly comes down to a chance discovery. However, a common trick is to differentiate something multiplied by a function $e^{g(x)}$, get some sort of expression, and then cancel out the original $e^{g(x)}$ term (that stays with us during differentiation) to get some polynomial expression in $f$ and $f'$, or $g$ and $g'$, or whatever functions you are interested in.
A: You can simply apply Rolle's theorem to $h(x)=f(x)e^{-\lambda x}$, where $g(x)=-\lambda x$. All you have to do is check if $h(x)$ satisfies the conditions of the Rolle's theorem. 
$ h(a)=f(a)e^{-\lambda a}=0$ and $h(b)=f(b)e^{-\lambda b}=0$ therefore $h(a)=h(b)$. 
$e^{-\lambda x}$ is differentiable over $\mathcal{R}$. We are given that
$f$ is continuous on $[a,b]$ and $f$ is differentiable on $(a,b)$.
Now by Rolle's theorem there exists a $c$ in $[a,b]$ such that
$$
h'(c)=0\rightarrow f'(c)=\lambda f(c) \mbox{ as desired}
$$
A: Let $h(x) = f(x)e^{g(x)}$.
Then we know by Rolle's theorem that $\exists c$ s.t. $h^{\prime} = f^{\prime} (c)e^{g(c)} +  f(c) g^{\prime}(c) e^{g(c)} = 0$.
Eliminating comon factors and rearranging, we get the $f^{\prime}(c) = -g^{\prime}(c)f(c)$
we want $-g^{\prime}(x) = \lambda$, so $g(x) = -\lambda x + k$ for some constant $k$ is the answer. 
