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

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.

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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. 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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