Question

The function $f\left(x\right)=\{\begin{array}{ccc}{x}^4+x^2-x+2& for& x\le 1\\ 3x^3-x^2+x& for& x>1\end{array}$ is

• A. continuous everywhere
• B. differentiable everywhere
• C. differentiable at x=1
• D. such that f ' exists everywhere but f '' is not continuous at x=1

Answer 1

The correct option is A continuous everywhere

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(x^4+x^2-x+2)=1+1-1+2$

$=3$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(3x^3-x^2+x)=3-1+1$

$=3$

$f\left(1\right)=1+1-1+2=3$

RHL $=f\left(1\right)=$LHL

$\therefore$ continuous everywhere

$f^{\prime }\left(x\right)=\{\begin{array}{c}4x^3+2x-1forx\le 1\\ 9x^2-2x+1forx>1\end{array}$

$\underset{x\to 1^{-}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 1^{-}}{lim}(4x^3+2x-1)=4+2-1$

$=5$

$\underset{x\to 1^{+}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 1^{+}}{lim}(9x^2-2x+1)=9-2+1$

$=8$

RHL$\ne$LHL

$\Rightarrow f^{\prime }\left(x\right)$is not differentiable at $x=1$