Question

The function $f(x) = |x -3| , x \ge 1 = \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4} , x < 1 $ , is

• A. not continuous at x = 1
• B. not derivable at x = 1
• C. continuous and derivable at x = 1
• D. continuous at x = 1 but not derivable at x = 1.

Answer 1

Answer: (C)

$Lt_{x\to1-}f\left(x\right) =Lt_{x\to1} \left(\frac{x^{2}}{4}- \frac{3x}{2}+\frac{13}{4}\right) $
$= \frac{1}{4} - \frac{3}{2} + \frac{13}{4} = 2$
$ Lt_{x\to1+} f\left(x\right) = Lt_{x\to1} \left(x-3\right) = 2 $
$\therefore Lt_{x \to1} f\left(x\right)=2 =f\left(1\right)$
$ \therefore f\left(x\right) $ is continuous at x=1
Again $Lf'\left(1\right) =Lt_{x\to1-} \frac{f\left(x\right)-f\left(1\right)}{x-1}$
$ = Lt_{x\to1} \frac{\frac{x^{2}}{4} - \frac{3x}{2} + \frac{13}{4} -2}{x-1} = Lt_{x\to1} \frac{x^{2} -6x-5}{4\left(x-1\right)} $
$=Lt_{x\to1} \frac{\left(x-5\right)}{4} = \frac{1-5}{4} = -1 $
$Rf'\left(1\right) =Lt_{x\to1+} \frac{f\left(x\right)-f\left(1\right)}{x-1} $
$=Lt_{x\to1} \frac{3-x-2}{x-1} =-1$
$ \therefore f'\left(1\right) =-1$
$ \therefore f'\left(x\right)$ is derivable at x = 1