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

MATH110 April 2016

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Question 09 (b)
Suppose that the function $f$ is differentiable for all $x$ and that $f(-1)=-1$ and $f(2)=5$. (b) Suppose also that $f''<0$ everywhere. Is it possible that $f'(2)=3$? Explain why or why not.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Note that ${\frac {f(2)-f(-1)}{2-(-1)}}=2$, and try to imagine a function that does satisfy $f''<0$ everywhere and $f'(2)=3$. To determine whether such a function exists, sketch an example of a differentiable function that satisfies $f(-1)=-1$, $f(2)=2$, and $f'(2)=3$, then try to use the Mean Value Theorem (as many times as you need to) to explain why your drawings end up the way they do.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.