Science:Math Exam Resources/Courses/MATH110/December 2012/Question 06

MATH110 December 2012

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[hide] Question 06
Let $f(x)={\begin{cases}{\sqrt {x-{\frac {3}{4}}}}&x\leq 1\\ax+b&x>1\end{cases}}$ Find constants $a$ and $b$ such that the function $f(x)$ is differentiable on the interval $(-\infty ,\infty )$ . Justify your answer.

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[show] Hint 1
In order for a function to be differentiable over the domain, it should be differentiable within each piece of the function as well as on the boundary. What are the conditions for a function to be differentiable at a point? How is differentiability related to continuity?

[show] Hint 2
In order for a function to be differentiable at a point, the derivative must be equal on both sides of the point. Also, differentiable functions are also continuous, meaning that the value of the function must also be equal on both sides of any point.

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[show] Solution
In order for $f(x)$ to be differentiable, the derivative at any point must be equal from both the left and right. Since $f(x)$ is clearly differentiable at all points in $(-\infty ,\infty )$ other than x = 1, our point of interest is $x=1$ . The derivative of $f(x)$ at x = 1 from the left will be $\left({\sqrt {x-{\frac {3}{4}}}}\right)'={\frac {1}{2{\sqrt {x-{\frac {3}{4}}}}}}$ which when evaluated at x = 1 gives ${\frac {1}{2{\sqrt {1/4}}}}=1$ The derivative of $f(x)$ at x = 1 from the right will be $(ax+b)'=a$ To be differentiable, these must be equal, so $a=1$ This gives us the value for a. Now we also know that in order for the function to be differentiable, it must also be continuous. In particular, this means both the right and left hand limits at x = 1 must be equal. So we have: ${\begin{aligned}\displaystyle \lim _{x\to 1^{-}}f(x)&=\lim _{x\to 1^{+}}f(x)\\\displaystyle \lim _{x\to 1^{-}}{\sqrt {x-{\frac {3}{4}}}}&=\lim _{x\to 1^{+}}1x+b\\\displaystyle {\sqrt {1/4}}&=1+b\\{\frac {-1}{2}}&=b\end{aligned}}$ So $f(x)$ will be differentiable when a = 1 and b = -1/2