Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^3-3x+2,& x<2\\ x^3-6x^2+9x+2,& x\ge 2\end{array}$ Then

• A. $\underset{x\to 2}{lim}f\left(x\right)$ does not exist
• B. f is not continuous at $x=2$
• C. f is continuous but not differentiable at $x=2$
• D. f is continuous and differentiable at $x=2$

Answer 1

Answer: (C)

f is continuous but not differentiable at $x=2$

$\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}(x^3-3x+2)$

$=2^3-3\left(2\right)+2$

$=4$

$\underset{x\to 2}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(x^3-6x^2+9x+2)$

$=2^3-6{\left(2\right)}^2+9\left(2\right)+2$

$=4$

LHL$=$RHL $\Rightarrow$continuous at $x=2$

$f^{\prime }\left(x\right)=\{\begin{array}{c}3x^2-3x<2\\ 3x^2-12x+9x\ge 2\end{array}$

$\underset{x\to 2^{-}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 2^{-}}{lim}(3x^2-3)$$=3{\left(2\right)}^2-3=9$

$\underset{x\to 2^{+}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 2^{+}}{lim}(3x^2-12x+9)$$=3{\left(2\right)}^2-12\left(2\right)+9$$=-3$

LHL$\ne$RHL $\Rightarrow$not differentiable at $x=2$