I have been asked to prove the following question for my Real Analysis course, and I'm not quite sure how to go about proving this. The textbook is Spivak's Calculus, and this is question 10 from Chapter 7. Thank you in advance! Suppose $f,g$ are continuous on $[a, b]$ and that $f(a) < g(a)$, bu...

Suppose $f$ and $g$ are continuous on $[a,b]$ and that $f(a) < g(a)$ but $f(b) > g(b)$. Prove that $f(x) = g(x)$ for some $x \in [a,b]$.

If $f$ and $g$ are continuous on $[a,b]$, $f(a)$ $\le$ $g(a)$, and $g(b)$ $\le$ $f(b)$, prove there is a point $c$ $\in [a,b]$ such that $f(c)$ $=$ $g(c)$. Any ideas on how to solve? I think I have to use the Intermediate Value Thm but I'm not sure.

If $f$ and $g$ are two continuous functions from $[a,b]$ to real numbers s.t $f(a)\lt g(a)$ and $f(b)\gt g(b)$. Prove that $f(x)=g(x)$ has at least 1 solution $c$ in $(a,b)$. I tried using intermediate value theorem. It states that if f continuous on [a,b] f(a)< d< f(b) then there exists a point...

First of all, let me write the statement properly: Theorem : Let $f(x)$ and $g(x)$ are continuous on a closed interval $[a,b]$. If $f(a)< g(a)$ and $f(b)>g(b)$, then there exists a $c$ in the interval $[a,b]$ such that $f(c)=g(c)$. I am new at proofs, so I wanted ask if the proof below c...

Suppose $f,g: [a,b] \to \mathbb{R}$ are continuous such that $f(a) < g(a)$ and $f(b) > g(b)$. Prove that there is $c \in [a,b]$ such that $f(c) = g(c).$ Attempt: Let $h(x) = f(x) - g(x)$. Then $h(a) < 0$ and $h(b) > 0$ while there exists a $c$ such that $h(c) = 0$. Then isn't it obvious tha...