$f:[1,4]\rightarrow[7,14]$ is a strictly concave surjective function then prove that $(f'(x))^2=49/9$ has at least one and at most two roots in $[1,4]$
$\displaystyle f'(x)=\pm\frac{7}{3}$, so if we need to prove that a line with slope $ \displaystyle\pm\frac{7}{3}$ is tangent to the function at least once and not more than two times.
The graph is enclosed in a rectangle with coordinates $(1,7),(1,14),(4,7)$ and $(4,14)$
Case 1 (the function is concave increasing): Since it's surjective and increasing $(1,7)$ and $(4,14)$ lie on the function. We can prove $(f'(x))^2=49/9$ has one root by Lagranges mean value theorem (since the slope of the diagonal of the rectangle is $7/3$).
Case 2 (the function is concave decreasing): Same logic as above, only this time the diagonal with slope $-7/3$ will be a tangent.
Case 3 (function is first increasing, then becomes decreasing or starts as decreasing and becomes increasing): This also case seems correct when I draw a graph, but I can't think of a rigorous proof.
Can anyone solve this? It's fine if you use another method. Thanks
Best Answer
Concave functions attain the minimum on a compact interval at one of the boundary points, therefore $f(1) = 7$ or $f(4) = 7$. Let us assume that $f(1) = 7$, the other case works similarly.
If $f(4) = 14$ then (as you said) $f'(\xi) = 7/3$ for some $\xi \in (1, 4)$ by the mean-value theorem.
Otherwise $f(c) = 14$ for some $c \in (1, 4)$. Then consider the function $$ g(x) = f(x) - L(x) $$ where $L$ is the linear function connecting $(1, 7)$ with $(4, 14)$. Note that $L$ has constant derivative $7/3$. From $g(1) = 0$, $g(c) > 0$ and $g(4) < 0$ we can conclude that $g$ attains its maximum at a point $\xi \in (1, 4)$, and then $$ 0 = g'(\xi) = f'(\xi) - \frac 73 \, . $$
It remains to show that $(f'(x))^2=49/9$ cannot have three or more solutions. In that case, $f'(x) = 7/3$ or $f'(x) = -7/3$ has at least two solutions, and that is not possible because $f'$ is strictly decreasing if $f$ is strictly concave.