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)

$L{t}_{x\to 1-}f\left(x\right)=L{t}_{x\to 1}\left(\frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4}\right)$
$=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=2$
$L{t}_{x\to 1+}f\left(x\right)=L{t}_{x\to 1}\left(x-3\right)=2$
$\therefore L{t}_{x\to 1}f\left(x\right)=2=f\left(1\right)$
$\therefore f\left(x\right)$ is continuous at x=1
Again $L{f}^{\prime }\left(1\right)=L{t}_{x\to 1-}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$=L{t}_{x\to 1}\frac{\frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4}-2}{x-1}=L{t}_{x\to 1}\frac{x^2-6x-5}{4\left(x-1\right)}$
$=L{t}_{x\to 1}\frac{\left(x-5\right)}{4}=\frac{1-5}{4}=-1$
$R{f}^{\prime }\left(1\right)=L{t}_{x\to 1+}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$=L{t}_{x\to 1}\frac{3-x-2}{x-1}=-1$
$\therefore f^{\prime }\left(1\right)=-1$
$\therefore f^{\prime }\left(x\right)$ is derivable at x = 1