Question

Let f be the function defined as follows:

Answer 1

With $f$ defined by

$f(x)=\begin{cases}|x-1|+2&\text{for }x$

in order for it to be continuous at $x=c$, we require

$\displaystyle\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)$

(i) If $a=2$ and $b=3$, then $f(1)=2(1)^2+3(1)=5$ and

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}|x-1|+2=2$

$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}2x^2+3x=5$

The limits don't match, so $f$ is not continuous at $x=1$ under these conditions.

(ii) To establish continuity at $x=1$, we'd need the limit as $x\to1$ from the right to be equal to the limit from the left, or

$\displaystyle\lim_{x\to1}ax^2+bx=\lim_{x\to1}|x-1|+2\iff a+b=2$

(iii) We have $f(2)=0$ and

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2}ax^2+bx=4a+2b$

$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2}5x-10=0$

For $f$ to be continuous at $x=2$, then, we'd need to have

$4a+2b=0$

(iv) Taking both requirements from parts (ii) and (iii), we solve for $a,b$:

$\begin{cases}a+b=2\\4a+2b=0\end{cases}\implies a=-2,b=4$

I've attached a plot that confirms this is correct.