Science:Math Exam Resources/Courses/MATH110/December 2013/Question 10 (b)/Solution 1

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

$\underset{x\to 2^{+}}{\lim }f(x)=f(2)=\underset{x\to 2^{-}}{\lim }f(x)$

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

$\begin{array}{rl}\underset{x\to 2^{-}}{\lim }f(x)& =(2{)}^2-2(2)+1=1\\ \underset{x\to 2^{+}}{\lim }f(x)& =a(2)+b=2a+b\end{array}$

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

$\underset{x\to 2^{+}}{\lim }\frac{f(x)-f(2)}{x-2}=\underset{x\to 2^{-}}{\lim }\frac{f(x)-f(2)}{x-2}$

Calculating these limits we find

$\begin{array}{rl}\underset{x\to 2^{+}}{\lim }\frac{f(x)-f(2)}{x-2}& =\underset{x\to 2^{+}}{\lim }\frac{x^2-2x+1-1}{x-2}\\ & =\underset{x\to 2^{+}}{\lim }\frac{x(x-2)}{x-2}\\ & =\underset{x\to 2^{+}}{\lim }x\\ & =2\end{array}$

and using $b-1=-2a$ from above we find that the left-handed limit is

$\begin{array}{rl}\underset{x\to 2^{-}}{\lim }\frac{f(x)-f(2)}{x-2}& =\underset{x\to 2^{-}}{\lim }\frac{ax+b-1}{x-2}\\ & =\underset{x\to 2^{-}}{\lim }\frac{ax+(-2a)}{x-2}\\ & =a\underset{x\to 2^{-}}{\lim }\frac{x-2}{x-2}\\ & =a\underset{x\to 2^{-}}{\lim }1\\ & =a\end{array}$

For the previous two limits to be equal we require that $a=2$. 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 $\frac{d}{dx}(ax+b)=a$.