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 $f'$ are differentiable everywhere and we can use the Mean Value Theorem for $f$ and $f'$ 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 $f'(s)={\frac {f(b)-f(a)}{b-a}}={\frac {f(2)-f(-1)}{2-(-1)}}={\frac {5-(-1)}{2-(-1)}}={\frac {6}{3}}=2$.

Now let $a=s$ and $b=2$, where $s$ is as defined above. By the Mean Value Theorem applied to $f'$ on $[a,b]$, there is a number $t$ such that $s=a<t<b=2$ and $f''(t)={\frac {f'(b)-f'(a)}{b-a}}={\frac {f'(2)-f'(s)}{2-s}}={\frac {3-2}{2-s}}={\frac {1}{2-s}}>0$. But then, $f''(t)>0$ contradicting the assumption that $f''<0$ everywhere. Hence, no such function $f$ exists.