Science:Math Exam Resources/Courses/MATH104/December 2013/Question 01 (k)

MATH104 December 2013

• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q1 (l) • Q1 (m) • Q1 (n) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q2 (e) • Q2 (f) • Q2 (g) • Q3 • Q4 • Q5 • Q6 (a) • Q6 (b) • Q6 (c) •

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[hide] Question 01 (k)
At $x=0,$ which of the following is true for the function $f(x)=x^{2}+e^{-2x}$ ? (A) $f$ is increasing. (B) $f$ is decreasing. (C) $f$ is discontinuous. (D) $f$ has a local minimum. (E) $f$ has a local maximum.

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.

[show] Hint 1
In order to find maxima and minima (and intervals of increasing and decreasing), we need to start taking derivatives.

[show] Hint 2
If the derivative exists everywhere - we can eliminate one of the incorrect answers above.

[show] Hint 3
Lastly, if we have a maximum or minimum at $\displaystyle x=0$ , what value should the derivative be at this point? What value would the derivative be if the function is increasing or decreasing at $\displaystyle x=0$ (this relates to the sign of the derivative)?

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.
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[show] Solution
We are given that $\displaystyle f(x)=x^{2}+e^{-2x}$ . Taking the derivative yields $\displaystyle f'(x)=2x-2e^{-2x}$ . Since this exists everywhere, our function is differentiable hence continuous thus (C) is not the correct answer. Factoring gives $\displaystyle f'(x)=2(x-e^{-2x})$ When we plug in $\displaystyle x=0$ , we see that the $\displaystyle f'(0)=2(0-e^{-2(0)})=-2$ . As this is negative, we know that $\displaystyle f(x)$ is decreasing at $\displaystyle x=0$ . Thus the correct answer is (B).