Science:Math Exam Resources/Courses/MATH110/December 2011/Question 06

MATH110 December 2011

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Question 06
Let $f(x)={\begin{cases}{\frac {1}{2}}\cos(x)&{\hbox{if }}x<a\\x^{2}+b&{\hbox{if }}x\geq a\end{cases}}$ Find constants a and b such that ƒ is differentiable everywhere. Justify your answer.

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
For a function to be differentiable at a point x = a, it must be continuous at a and the derivative on both sides of a must be equal.

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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. In order for $f(x)$ to be differentiable at a, we have two requirements: that the function be continous and derivatives from the right and left of a point are the same. Before we start, let us note that ƒ is surely, as a composition of differentiable functions, differentiable at all points expect x = a. Hence we can focus our study on the point x = a. We begin with the second condition: that the derivative from the right ($x<a$) must be the same as the derivative from the left ($x>a$). This means that we are trying to find the value a such that $\displaystyle -{\frac {1}{2}}\sin a=2a$ Simplifying a bit, this is the same as $\displaystyle -\sin a=4a$ If we think about the function $-\sin x$ we know that it is bounded between -1 and 1. If a was greater than 1/4, say, 1/2, the left side would be less than one, while the right side would already be greater than 2. So a must be between -1/4 and 1/4 for this equation to hold. In fact, the only place that the graphs of $-\sin x$ and 4x intersect is at x = 0. So a = 0. This satisfies the condition that the derivatives match; now we must ensure that $f(x)$ is continuous. The function is defined everywhere, so we must check that the limit at x = a exists. The limit from the left of x = 0 is $\displaystyle \lim _{x\to 0^{-}}f(x)=\lim _{x\to 0^{-}}{\frac {1}{2}}\cos x={\frac {1}{2}}$ The limit from the right of x = 0 is $\displaystyle \lim _{x\to 0^{+}}f(x)=\lim _{x\to 0^{+}}x^{2}+b=b$ These two limits must be equal, so we have b = 1/2.

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