Show that $\exists c\in (0,1)$ such that $e^{f'(c)}=f(c)+1.$ 
Question: Let $f:[0,1]\in\mathbb{R}$ be a continuous function satisfying $\int_0^1f(x)dx=0$ and $f(0)f(1)>0$. Assume further that $f'$ is continuous on $(0,1)$. Show that $\exists c\in (0,1)$ such that $$e^{f'(c)}=f(c)+1.$$

My solution: It is given that $f(0)f(1)>0\implies f(0)>0,f(1)>0$ or $f(0)<0, f(1)<0$. Let us assume WLOG, that $f(0)>0,f(1)>0$. 
Now since $\int_0^1f(x)dx=0$ and $f(0)>0,f(1)>0$, thus we must have a point $a\in (0,1)$ such that $f(a)<0$. Now since $f$ is continuous on $[0,1]$, therefore by IVT we can conclude that $\exists c_1\in (0,a)$, such that $f(c_1)=0$ and $\exists c_2\in(a,1),$ such that $f(c_2)=0$. From this we can conclude, there exists at least two distinct roots of $f$ in $(0,1)$. Therefore we are sure to have a point $b,0<b<1$ such that $f(x)>0, \forall x\in (0,b)$ and $f(b)=0$. Also we are sure to have a point $a,0<a<e<1$, such that $f(x)<0, \forall x\in(a,e)$ and $f(e)=0$. 
Now consider the interval $[0,b]$. We have $f(0)>0$ and $f(b)=0$. Thus by MVT, we can conclude that $\exists c_3\in (0,b)$, such that $$f'(c_3)=\frac{f(b)-f(0)}{b-0}=-\frac{f(0)}{b}<0.$$ Now since $c_3\in (0,b)\implies f(c_3)>0.$
Again consider the interval $[a,e]$. We have $f(a)<0$ and $f(e)=0$. Thus by MVT, we can conclude that $\exists c_4\in (a,e)$, such that $$f'(c_4)=\frac{f(e)-f(a)}{e-a}=-\frac{f(a)}{e-a}>0.$$ Now since $c_4\in (a,e)\implies f(c_4)<0.$
Now let $g(x):=e^{f'(x)}-f(x)-1, \forall x\in(0,1).$ Observe that $g$ is continuous on $(0,1)$. 
We have $g(c_3)=e^{f'(c_3)}-f(c_3)-1.$ 
Now $f'(c_3)<0\implies e^{f'(c_3)}<1$ and $f(c_3)>0\implies f(c_3)+1>1\implies -f(c_3)-1<-1.$ 
Thus $e^{f'(c_3)}-f(c_3)-1<0\implies g(c_3)<0.$
A similar reasoning shows that $g(c_4)>0$. 
Thus by IVT, we can conclude that $\exists c\in(c_3,c_4)\subseteq(0,1)$, such that $g(c)=0\implies e^{f'(c)}=f(c)+1.$
Does this solution works fine? And is there a more better solution?
 A: WLOG, we suppose $f(0)>0,f(1)>0$. Define
$$F(x):=e^{f'(x)}-f(x)-1, \forall x\in(0,1).$$
On the one hand,
$\int_0^1f(x)dx=0$ 
implies $f$ can attain its Minimum value at some point $x_0\in(0,1)$,
such that $f(x_0)<0,$ and $f'(x_0)=0$(Fermat's lemma).
So
$$F(x_0)=-f(x_0)>0.$$
On the other hand, $f(0)>0,f(x_0)<0$ implies that: the zeros set
$$\{x\mid x\in[0, x_0], f(x) =0\}\ne \emptyset.$$
(by intermediate value theorem)
Thanks  @ Paramanand Singh for pointing out that: the zeros set
$$\{x\mid x\in[0, x_0], f(x) =0\}$$
has  minimum  and maximum element no matter it is finite set and infinite set.(Proof can be found Intermediate value theorem and supremum)
Let $$x_1=\max\{x\mid x\in[0, x_0], f(x) =0\},$$
and obviously $x_1\in(0, x_0)$.
So $f(x_1)=0$ and $f(x)<0$ for $x\in(x_1,x_0)$ .
Consider the derivative $f'(x_1)$, we know that
$$f'(x_1)=\lim_{x\to x_1^+}\frac{f(x)-f(x_1)}{x-x_1}
=\lim_{x\to x_1^+}\frac{f(x)}{x-x_1}\leq 0.$$
If $f'(x_1)=0$, take $c=x_1$, we can get $e^{f'(c)}=f(c)+1$.
If $f'(x_1)<0$, then $F(x_1)=e^{f'(x_1)}-1<0$,
by intermediate value theorem, we can conclude that $\exists\  c\in(x_1,x_0)\subset(0,1)$,
such that
$$F(c)=0\iff e^{f'(c)}=f(c)+1.$$
A: If such $c$ doesn't exist, either always $e^{f'}>f+1$ or always $e^{f'}<f+1$. Whenever $f=0$, the former means $f'>0$ always and the latter means $f'<0$ always. However $f$ must change sign at least twice, and this is impossible. 
