TITLE: How can the following be false based on the information about continuous functions?
QUESTION [8 upvotes]: Let $f\colon\mathbb{R}\to [0,\infty)$ be a continuous function. Then, what is the rationale behind saying that the following are false statements:

There exists $x\in\mathbb{R}$ such that  $f(x)=\int_{-1}^{1}f(t)dt$.
There exists $x\in\mathbb{R}$ such that  $f(x)=\frac{f(0)+f(1)}{2}$.

I think the second statement should be true due to the intermediate value theorem but I am not certain about the first statement. But  my solution manual says both statements are false. What should be the rationale behind this reasoning?

REPLY [15 votes]: The first statement reminds of the Mean Value Theorem for Integrals, but… The Mean Value Theorem for Integrals says that if $f$ is continuous on $[a,b]$, then there exists $x\in[a,b]$ such that
$$f(x)=\frac{1}{b-a}\int_a^b f(t)\,dt.$$
In this case, $a=-1$ and $b=1$, so we can guarantee that there exists $x\in[-1,1]$ such that
$$f(x)=\frac{1}{2}\int_{-1}^1 f(t)\,dt.$$
This is very similar to the given equality, but without the $1/2$. And that suggests that statement (1) is probably false. After realizing that, it's not difficult to come up with a counterexample.
As for statement (2), it's actually true by the Intermediate Value Theorem, since $f$ is continuous.