Science:Math Exam Resources/Courses/MATH100/December 2010/Question 04 (d)

MATH100 December 2010

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Question 04 (d)

Full-Solution Problems. In questions 2-8, justify your answers and show all your work. If a box is provided, write your final answer there. Simplification of answers is not required unless explicitly requested. Let $f(x)={\begin{cases}{\frac {4}{\pi }}\tan ^{-1}x&{\text{if }}x\geq 1\\2-x^{4}&{\text{if }}x<1\end{cases}}$ NOTE: Another notation for tan-1(x) is arctan(x) Determine open intervals where the graph of ƒ is concave upwards or concave downwards, and the x-coordinates of all inflection points (if any).

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
How does concavity relate to the derivatives of the function f(x)?

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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. To determine the intervals of concavity, we must evaluate ƒ''(x) and determine where it is positive and negative: $f(x)={\begin{cases}{\frac {4}{\pi }}\tan ^{-1}(x)&{\text{if }}x\geq 1\\2-x^{4}&{\text{if }}x<1\end{cases}}$ Evaluating the second derivative of f gives $f''(x)={\begin{cases}{\frac {4}{\pi }}{\frac {-2x}{(1+x^{2})^{2}}}&{\text{if }}x>1\\-12x^{2}&{\text{if }}x<1\end{cases}}$ From this we see that the possible inflection points are x = 0. We need only confirm that the sign of the concavity changes across x = 0. We must also consider concavity changes across x = 1 since ƒ is not differentiable there. When x < 1, the second derivative is always negative since -12x2 is always negative. When x > 1, the second derivative is always negative as well since in that case, -2x is always negative. From these results we can see that ƒ(x) is concave down for all x except at the critical points. In other words, ƒ(x) is concave down on $(-\infty ,0),(0,1),(1,\infty )$ and so, there are no inflection points.

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