Science:Math Exam Resources/Courses/MATH104/December 2016/Question 08 (a)

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MATH104 December 2016

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Question 08 (a)
The graph on the left is the derivative of $y=f'(x)$. Given that $f(0)=0$, sketch the graph of $f(x)$ on the right grid. Clearly lable any local extrema and inflection points. graphs

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
From the graph of $y=f'(x)$, determine where $f$ is increasing or decreasing and concave up or down.

Hint 2
Find the points satisfying $f'(x)=0$ and $f''(x)=0$, which are the candidates of the local extrema and inflection points, respectively.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

From the graph of $y=f'(x)$, we can first see that $f'(2)=f'(4)=0$, so that both are critical points; the candidates of the local extrema. Since $f'>0$ on $(0,2)\cup (4,5)$, while $f'<0$ on $(2,4)$, $f$ is increasing on $(0,2)\cup (4,5)$ and decreasing on $(2,4)$. This implies that $x=2$ is a local maximum and $x=4$ is a local minimum. On the other hand, we observe that $f''(1)=f''(3)=0$, and hence both are the candidates of inflection points. Indeed, this follows from that $x=1$ and $x=3$ are local extrema of $f'$ and hence the critical points of $f'$. Finally, the graph tells us that $f'$ increases on the interval $(0,1)$ and $(3,5)$, while it decreases on $(1,3)$. This determines the sign of the second derivative of $f$ and hence the concavity of the function $f$ as follows; Intervals $(0,1)$ $(1,3)$ $(3,5)$ Sign of $f''$ $+$ $-$ $+$ Concavity Up Down Up Since the sign of the second derivative is changing at both points $x=1$ and $x=3$, they are inflection points of $f$. To summarize, we obtain a local maximum $x=2$ and a local minimum $x=4$ and two inflection points $x=1,3$. Then, based on our observation, the graph of $f$ can be sketched as follows; graphs

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