MATH110 April 2019
Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q2 (a)(i) • Q2 (a)(ii) • Q2 (a)(iii) • Q2 (b) • Q2 (c) • Q3 • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 • Q7 (a) • Q7 (b) • Q7 (c) • Q8 • Q9 • Q10 (a) • Q10 (b) •
[hide]Question 05 (c)
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Let .
Find the interval(s) on which f(x) is concave down.
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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?
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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.
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[show]Hint 1
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Recall that a function f(x) is concave down at all points where its second derivative is negative; that is, f ′′(x) < 0.
Computing f ′(x) is straightforward, but computing f ′′(x) will require more effort. The final expression is not that messy!
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[show]Hint 2
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After simplification, the second derivative is:
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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.
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[show]Solution
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Recall that a function f(x) is concave down at all points where its second derivative is negative; that is, f ′′(x) < 0.
From part (a), we know that
Using the quotient rule again, we find that
This simplifies to
The denominator is always positive (by the same reasoning as in the solution to part (b)), so f "(x) will be negative whenever the numerator is negative, in symbols,
To figure out for which values of x the inequality holds, factor the expression as
We do this because it's easier to figure out for what x each of , or is negative and we know that a product of three numbers is negative if one of them is negative and the other two are positive or if all three are negative. Therefore if or if is in the interval , in which case and is positive.
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