Science:Math Exam Resources/Courses/MATH100/December 2019/Question 7

MATH100 December 2019

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Question 7
Find, with proof, the real numbers $a$ and $b$ which make the following function differentiable at $x=0$. $f(x)=\sin(bx)+2,x\leq 0$ and $f(x)=ax+b,x>0$.

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
Think about what it means for the function $f(x)$ to be differentiable at a point in its domain. Which limits must you evaluate (yes, this will require more than one limit)? And how does this connect to imposing a condition on the real numbers $a,b$?

Hint 2

Remember, in order for a function $f(x)$ to to be differentiable at a real number $x_{0}$ in its domain, it must satisfy the following conditions: (1.) $\lim _{h\rightarrow 0^{+}}{\frac {f(x_{0}+h)-f(x_{0})}{h}}=\lim _{h\rightarrow 0^{-}}{\frac {f(x_{0}+h)-f(x_{0})}{h}}=L,$ for some real number $L$. (2.) $\lim _{x\rightarrow x_{0}^{+}}f(x)=\lim _{x\rightarrow x_{0}^{-}}f(x)=M,$ for some real number $M$. If both conditions are satisfied, we say that $f$ is differentiable at $x_{0}$, with $f^{'}(x_{0})=L$.

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

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 the piecewise defined function $f(x)$ to to be differentiable at $0$, it must satisfy the following conditions: (1.) $\lim _{h\rightarrow 0^{+}}{\frac {f(0+h)-f(0)}{h}}=\lim _{h\rightarrow 0^{-}}{\frac {f(0+h)-f(0)}{h}}=L,$ for some real number $L$. (2.) $\lim _{x\rightarrow 0^{+}}f(x)=\lim _{x\rightarrow 0^{-}}f(x)=M,$ for some real number $M$. Let us evaluate the first condition, for arbitrary real numbers $a,b$. First, we know: $\lim _{h\rightarrow 0^{+}}{\frac {f(0+h)-f(0)}{h}}=\lim _{h\rightarrow 0^{+}}{\frac {(ah+b)-b}{h}}=\lim _{h\rightarrow 0^{+}}a=a$, $\lim _{h\rightarrow 0^{-}}{\frac {f(0+h)-f(0)}{h}}=\lim _{h\rightarrow 0^{-}}{\frac {\sin(bh)+2-(\sin(0)+2)}{h}}=\lim _{h\rightarrow 0^{-}}{\frac {\sin(bh)}{h}}=\lim _{h\rightarrow 0^{-}}{\frac {b\cos(bh)}{1}}=b$. Here, in the left-hand limit (the second limit we evaluated), we utilized L'opital's rule. Hence, for these limits to be equal, we must set $a=b$. Now, let us check the second condition, for arbitrary real numbers $a,b$. We know: $\lim _{x\rightarrow 0^{+}}f(x)=\lim _{x\rightarrow 0^{+}}ax+b=b$ $\lim _{x\rightarrow 0^{-}}f(x)=\lim _{x\rightarrow 0^{-}}\sin(bx)+2=2$. Therefore, for these two limits to be equal, we must set $b=2$. Finally, in summary, in order for our piecewise defined function $f(x)$ to be differentiable at $x_{0}=0$, we must set $a=b=2$. Making this choice, we see that $f^{'}(0)=b=2$.