Proving the equation $e^{x}/4 = x$ has exactly two solutions. I need to prove that $e^x/4 = x$ has exactly two solutions to it on the interval $[0, +\infty)$. I'm pretty sure that I will need to use the Intermediate Value Theorem.
Intermediate Value Theorem: Suppose $f : [a, b] \rightarrow \mathbb{R}$ is continuous. Let $c$ be a number that lies strictly between $f(a)$ and $f(b$). Then there exists an $x_{0} \in [a, b]$ for which $f(x_{0}) = c$.
My attempt (wrong):
Let $f(x) = e^x/4$. Clearly $f(x)$ is continuous for positive $x$. Thus, we can apply the Intermediate Value Theorem. Note that $f(0) = 1/4$ and $f(2) > 4$. Since the number $2$ lies strictly between $f(0)$ and $f(2)$, there is some $x_{0} \in [a, b]$ for which $f(x_{0}) = 2$. 
 A: Your attempt answers about the existence of at least a solution of $f(x)=4$, but doesn't count them.

Consider $f(x)=e^x-4x$. Since $f'(x)=e^x-4$, the function has a critical point at $x=\log 4$, whose value is $f(\log4)=4-4\log4<0$.
Finally,
$$
\lim_{x\to-\infty}f(x)=\lim_{x\to\infty}f(x)=\infty
$$
The intermediate value theorem provides for a single solution of $f(x)=0$ in $(-\infty,\log4)$ and a single solution in $(\log4,\infty)$ (uniqueness due to $f$ being respectively decreasing and increasing over those two intervals).

More generally, consider the equation $f(x)=0$, where $f(x)=e^x-kx$. If $k>0$, the pattern is the same:
$$
\lim_{x\to-\infty}f(x)=\lim_{x\to\infty}f(x)=\infty
$$
and $f'(x)=0$ for $x=\log k$. Since $f(\log k)=k(1-\log k)$, the equation will have solutions provided $1-\log k\le0$, that is, $k\ge e$. There will be a single solution for $k=e$ and two solutions for $k>e$. No solution for $0<k<e$.
For $k<0$, the derivative is $f'(x)=e^x-k>0$, but
$$
\lim_{x\to-\infty}f(x)=-\infty
\qquad
\lim_{x\to\infty}f(x)=\infty
$$
The equation has a single solution.
A: Note that $\frac{e^1}4<1$ and that $\frac{e^0}4=\frac14>0$. So, $\frac{e^x}4=x$ for some $x\in(0,1)$, by the intermediate value theorem. On the other hand, $\frac{e^4}4>\frac{2^4}4=4$ and so, again by the intermediate value theorem, $\frac{e^x}4=x$ for some $x\in(1,4)$.
Now, suppose that the equation $\frac{e^x}4=x$ has $3$ distinct solutions. Then, by the mean value theorem, the equation $\frac{e^x}4=1$ will have $2$ distinct solutions. But that's not possible, since $x\mapsto\frac{e^x}4$ is strictly increasing.
A: Study the sign variations of the function $e^x-4x$, which is continuous.
The derivative is $e^x-4$ and cancels once, at $x=\log 4$. 
Then
$$f(-\infty)>0,\\f(\log 4)<0,\\f(\infty)>0$$ proves exactly two real roots, as the function is monotonic in between.
