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

MATH110 December 2011

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[hide] 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.

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[show] 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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[show] Solution
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.