Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (b)/Solution 1

No such function exists. To see this, use the Mean Value Theorem twice. Let $a=-1$ and $b=2$. Because $f^{″}<0$ everywhere, $f$ and $f^{\prime }$ are differentiable everywhere and we can use the Mean Value Theorem for $f$ and $f^{\prime }$ on the closed interval $[a,b]$. By the Mean Value Theorem applied to $f$ on $[a,b]$, there is a number $s$ such that $-1=a<s<b=2$ and $f^{\prime }(s)=\frac{f(b)-f(a)}{b-a}=\frac{f(2)-f(-1)}{2-(-1)}=\frac{5-(-1)}{2-(-1)}=\frac{6}{3}=2$.

Now let $a=s$ and $b=2$, where $s$ is as defined above. By the Mean Value Theorem applied to $f^{\prime }$ on $[a,b]$, there is a number $t$ such that $s=a<t<b=2$ and $f^{″}(t)=\frac{f^{\prime }(b)-f^{\prime }(a)}{b-a}=\frac{f^{\prime }(2)-f^{\prime }(s)}{2-s}=\frac{3-2}{2-s}=\frac{1}{2-s}>0$. But then, $f^{″}(t)>0$ contradicting the assumption that $f^{″}<0$ everywhere. Hence, no such function $f$ exists.