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$f:[1,4]\rightarrow[7,14]$ 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

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2 Answers 2

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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.

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Since $f$ is surjective on the interval $[7,14]$ there exist $a,b$ in $[1,4]$ such that $f(a)=7$ and $f(b)=14$. Therefore, by MVT there is $c\in (1,4)$ such that $$|f'(c)|=\frac{14-7}{|b-a|}\geq \frac{14-7}{4-1}=\frac{7}{3}.$$ Moreover, again by MVT, there is $d\in (1,4)$ such that $$|f'(d)|=\frac{|f(4)-f(1)|}{4-1}\leq \frac{14-7}{4-1}=\frac{7}{3}.$$ Now by Darboux's theorem, $f'$ has the intermediate value property and therefore there is $t$ between $c$ and $d$ such that $|f'(t)|=7/3$, i.e. $(f'(t))^2=49/9$ for some $t\in (a,b)$.

As regards the number of solutions, note that $f(x)=7/3(x-1)+7$ is concave and $f'(x)=7/3$ identically!

On the other hand, if $f$ is strictly concave then $f'$ is strictly decreasing and therefore injective, which implies that $(f'(x))^2=49/9$ has at most two solutions: one for $f'(x)=7/3$ and another for $f'(x)=-7/3$.

Note that $f(x)=\frac{28}{9}(x-1)(4-x)+7$ satisfies the assumptions, it is strictly concave and $$(f'(x))^2=\left(\frac{28}{9}(5-2x)\right)^2=\frac{49}{9}$$ has exactly two solutions: $\frac{17}{8},\frac{23}{8}\in (1,4)$.

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