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

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

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 ${\frac {d}{dx}}(ax+b)=a$.