How to prove with rigor that there exists $c \in [-1,1]$ such that $f(c)=100c$?

Question

The function $f:[-1,1] \to \mathbb{R}$ is continuous on its domain, with $f(-1)=1$ and $f(1)=5$. The statement that $\exists c \in [-1,1]$ such that $f(c)=100c$ is intuitively true because a continuous graph starting from $(-1,1)$ to $(1,5)$ must cross the graph of $f(x)=100x$ at some point.

However, I am wondering how to prove it with rigor? Thanks.

Answer 1

Define $g(x) := f(x) - 100x$. This is a continuous function with the boundary values $g(-1)= 101$ and $g(1) = -95$. Due to the intermediate value theorem, there is a value $x_0\in (-1,1)$ such that $g(x_0) = 0$, or equivalently, $f(x_0) = 100 x_0$.

Answer 2

If you want to go lower than the intermediate value theorem: f(x) - 100x is continuous with f(-1) >= 0 and f(1) < 0. Let S be the set of x where f(x) >= 0. This set is non-empty with an upper bound, therefore due to the completeness axiom it has a least upper bound c. You can show that f(c) - 100c = 0 because if it was not 0 then f(x) -100x would have the same sign in an area around c.