# $f:[1,4]\rightarrow[7,14]$ is a concave surjective function then prove that $(f'(x))^2=49/9$ has at least one and at most two roots in $[1,4]$

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

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.

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: $$\frac{17}{8},\frac{23}{8}\in (1,4)$$.