Prove that there is at least one number from $[a;b] : f(c)=g(c)$

Question

$f(x)$ and $g(x)$ are two continuous functions at $[a;b]$ and for every number $x$ from $[a;b]$ there is at least one number $y$ from $[a;b] / f(x)=f(y)$

Prove that there is at least one number from $[a;b] : f(c)=g(c)$

Answer 1

I'm assuming that you meant $f(x) = g(y)$ not $f(x) = f(y)$ in the statement of the problem.

By our hypotheses, $f([a,b]) = [c,d] = g([a,b])$ for some $a,b$. Assume that $f(x_1) = c$ and $f(x_2) = d$ for some $x_1,x_2 \in [a,b]$. Then $f(x_1) \leq g(x_1)$ and $f(x_2) \geq g(x_2)$, and so the result follows by the Intermediate Value Theorem applied to the function $h(x) = f(x) - g(x)$ either on the interval $[x_1,x_2]$ if $x_1 < x_2$ or $[x_2,x_1]$ if $x_2 < x_1$.