Science:Math Exam Resources/Courses/MATH110/December 2013/Question 10 (b)

MATH110 December 2013

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Question 10 (b)
Find constants $a$ and $b$ so that the function $f(x)={\begin{cases}x^{2}-2x+1&{\text{if }}x\leq 2\\ax+b&{\text{if }}x>2\end{cases}}$ is differentiable everywhere.

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
Each branch of the function is differentiable (why?). Therefore, the only point where our function might not be differentiable is at the break point of the function (in this case, at $x=2$).

Hint 2
In order for a function to be differentiable at 2, it must also be continuous at 2. Use part (a) and the fact that we want continuity to help find these values.

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. As suggested in the hint, this function is continuous everywhere except for possibly at the point $x=2$. There we need to check continuity and differentiability there. For continuity, we require that the following three quantities are all the same $\lim _{x\to 2^{+}}f(x)=f(2)=\lim _{x\to 2^{-}}f(x)$ Since $f(2)=(2)^{2}-2(2)+1=1$ we have to check that both limits above also equal 1. Calculating the limits we find that ${\begin{aligned}\lim _{x\to 2^{-}}f(x)&=(2)^{2}-2(2)+1=1\\\lim _{x\to 2^{+}}f(x)&=a(2)+b=2a+b\end{aligned}}$ Thus we require $2a+b=1$. Rearranging this, we see that $b-1=-2a.$ The second piece of information come from checking that the function is differentiable. For differentiability, we require that the following two limits are the same $\lim _{x\to 2^{+}}{\frac {f(x)-f(2)}{x-2}}=\lim _{x\to 2^{-}}{\frac {f(x)-f(2)}{x-2}}$ Calculating these limits we find ${\begin{aligned}\lim _{x\to 2^{+}}{\frac {f(x)-f(2)}{x-2}}&=\lim _{x\to 2^{+}}{\frac {x^{2}-2x+1-1}{x-2}}\\&=\lim _{x\to 2^{+}}{\frac {x(x-2)}{x-2}}\\&=\lim _{x\to 2^{+}}x\\&=2\end{aligned}}$ and using $b-1=-2a$ from above we find that the left-handed limit is ${\begin{aligned}\lim _{x\to 2^{-}}{\frac {f(x)-f(2)}{x-2}}&=\lim _{x\to 2^{-}}{\frac {ax+b-1}{x-2}}\\&=\lim _{x\to 2^{-}}{\frac {ax+(-2a)}{x-2}}\\&=a\lim _{x\to 2^{-}}{\frac {x-2}{x-2}}\\&=a\lim _{x\to 2^{-}}1\\&=a\end{aligned}}$ For the previous two limits to be equal we require that $\displaystyle a=2$. Substituting back into $2a+b=1$ shows us that $b=-3$. Note. We could also have taken a bit of a short cut and seen that this function is differentiable if the derivatives of the two halves are equal at 2, that is ${\frac {d}{dx}}(x^{2}-2x+1)=2x-2$ at $x=2$ has to equal ${\frac {d}{dx}}(ax+b)=a$.

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