Suppose that $f$ is differentiable on $\mathbb{R}$, that $f(1)=1$ and $f(7)=5$. Show that there exists $c\in(0,4)$ such that $f'(c)=\frac{2}{3}$.

Question

So, it's obvious that using the Mean Value Theorem you can ascertain that there exists $\alpha\in(1,7)$ such that $$f'(\alpha)=\frac{f(7)-f(1)}{7-1}=\frac{4}{6}=\frac{2}{3}.$$ What I don't understand is how I'm meant to narrow the interval of $\alpha$ to $(0,4)$ from $(1,7)$. Any help would be appreciated.

Available content: Everything in Spivak's Calculus up to and including derivatives and their applications. You aren't meant to use integrals or anything more advanced for this.

Answer 1

This is not true, take for exemple $f=1$ on $]-\infty,6]$ and then connect $(6,1)$ and $(7,5)$ in a smooth way.