Calculus IVT question

Question

Let $f$ be a function that is continuous on $[a,b]$ for some $a<b$. Suppose that f(a)=f(b)=0, and let $M=\sup${$f(x) : x \in [a,b]$}, while $M>0$, and let $0 \leq r < M$.

Prove that the set {$x \in [a,b] : f(x)=r$} contains at least two distinct points.

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So, because $M>0$ and $M=\sup${$f(x) : x \in [a,b]$}, there exists $x_{0} \in [a,b]$ such that $f(x_{0})=M>0$, so we get $f(a) \leq r < f(x_{0})$.

$f$ is continuous on $[a,b]$ for $a<b$, so from IVT, there exists $c,r \in R$, $f(a)\leq r<f(x_{0})$ such that $f(c)=r$.

From here I am a bit confused on what to do next.

Any tip will be very appreciated!

Thanks a lot!

Answer 1

hint

Assume $ r>0$.

$ f $ is continuous at the compact $ [a,b] \implies $

$$(\exists c\in(a,b) )\;:\;f(c)=M$$

let $$g(x)=f(x)-r$$ $ g $ is continuous at $ [a,c] $ and $ [c,b]$. $$g(a)g(c)=-r(M-r)<0$$

and

$$g(c)g(b)=-r(M-r)<0$$

conclude by IVT.

Answer 2

You are almost there! Here I present the reasoning in its entirety.

Since $f$ is continuous defined on the compact interval $[a,b]$, it attains its maximum $f(x_{\max}) = M > 0$ for some $x_{\max}\in(a,b)$.

Consequently, once we have that $f(a) \leq r < f(x_{\max})$ and $f(b) \leq r < f(x_{\max})$, we conclude through the IVT there exists $x_{1}\in[a,x_{\max}]$ as well as $x_{2}\in[x_{\max},b]$ such that $f(x_{1}) = f(x_{2}) = r$, and we are done.

Hopefully this helps!