Science:Math Exam Resources/Courses/MATH100/December 2011/Question 07

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MATH100 December 2011

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Question 07
Full-Solution Problems. In questions 2-8, justify your answers and show all your work. Simplification of answers is not required unless explicitly stated. Let $f(x)={\begin{cases}4x-{\frac {\sin ^{2}(2x-2)}{3x-3}}-1,&{\textrm {if}}\;x<1,\\2x+1,&{\textrm {if}}\;x\geq 1.\end{cases}}$ Use the definition of the derivative to find ƒ’(1) or show that ƒ’(1) does not exist.

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.

Hint 1
Using the definition of a derivative, we see that $f'(1)=\lim _{x\to 1}{\frac {f(x)-f(1)}{x-1}}$

Hint 2
Since the function is defined piecewise and the function is broken up at 1, we will need to use the left and right hand limits to evaluate the derivative.

Hint 3
Remember that $\lim _{x\rightarrow 0}{\frac {\sin(x)}{x}}=1$ You will need a variant of this.

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. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Using the definition of a derivative, we see that $f'(1)=\lim _{x\to 1}{\frac {f(x)-f(1)}{x-1}}$ Since the functions differ on either side of 1, we will want to use one sided limit to attempt to evaluate the limit. If the one sided limits are equal, then the derivative exists. Otherwise, the derivative does not exist. Firstly, notice that $f(1)=2(1)+1=3$. From the left, we have ${\begin{aligned}\lim _{x\to 1^{-}}{\frac {f(x)-f(1)}{x-1}}&=\lim _{x\to 1^{-}}{\frac {4x-{\frac {\sin ^{2}(2x-2)}{3x-3}}-1-3}{x-1}}\\&=\lim _{x\to 1^{-}}{\frac {4x-4}{x-1}}-\lim _{x\to 1^{-}}{\frac {\frac {\sin ^{2}(2x-2)}{3(x-1)}}{x-1}}\\&=\lim _{x\to 1^{-}}{\frac {4(x-1)}{x-1}}-\lim _{x\to 1^{-}}{\frac {\sin ^{2}(2x-2)}{3(x-1)^{2}}}\\&=4-\lim _{x\to 1^{-}}{\frac {\sin ^{2}(2x-2)}{3(x-1)^{2}}}\end{aligned}}$ Now, to evaluate the last limit, recall that $\lim _{x\to 0}{\frac {\sin(x)}{x}}=1$ hence $\lim _{x\to 1}{\frac {\sin(2x-2)}{2x-2}}=1$ and so, multiplying top and bottom by 4, we have ${\begin{aligned}\lim _{x\to 1^{-}}{\frac {f(x)-f(1)}{x-1}}&=4-\lim _{x\to 1^{-}}{\frac {\sin ^{2}(2x-2)}{3(x-1)^{2}}}\\&=4-\lim _{x\to 1^{-}}{\frac {4\sin ^{2}(2x-2)}{3\cdot 4(x-1)^{2}}}\\&=4-\lim _{x\to 1^{-}}{\frac {4\sin ^{2}(2x-2)}{3\cdot (2x-2)^{2}}}\\&=4-{\frac {4}{3}}\left(\lim _{x\to 1^{-}}{\frac {\sin(2x-2)}{2x-2}}\right)^{2}\\&=4-{\frac {4}{3}}\\&={\frac {8}{3}}\end{aligned}}$ As for the other side, we have ${\begin{aligned}\lim _{x\to 1^{+}}{\frac {f(x)-f(1)}{x-1}}&=\lim _{x\to 1^{+}}{\frac {2x+1-3}{x-1}}\\&=\lim _{x\to 1^{+}}{\frac {2x-2}{x-1}}\\&=2\\\end{aligned}}$ As the left and right hand limits do not agree, we have that the derivative does not exist at 1.

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