Let f and g be differentiable functions on an interval $(a,b)$ such that $a < b$. If there is $x0 ∈ (a,b)$ for which $f(x0) =g(x0)$ and $f(x) ≤ g(x)$ for all $x ∈ (a,b)$, prove that $f’(x0) =g’(x0)$
I applied the MVT on f and g and I got: $∃x0 ∈ (a,b)$ : $f’(x0) = [f(a)-f(b)]/[a-b]$ $∃x0 ∈ (a,b)$ : $g’(x0) = [g(a)-g(b)]/[a-b]$ Thus: $f’(x0)/g’(x0)$ = $[f(a)-f(b)]/[g(a)-g(b)]$=... If I can prove that that’s equal to $f(x0)/g(x0) = 1$ Then I’m done
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$\begingroup$HINT: The function $h:=g-f$ has a minimum at $x_0$, what does it say about its derivative there?
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