Limits Continuity And Differentiability Question 8
Question: If $ f(x)=\sqrt[3]{\frac{x^{4}}{| x |}},x\ne 0 $ and f(0) = 0 is:
Options:
A) Continuous for all x but not differentiable for any x
B) Continuous and differentiable for all x
C) Continuous for all x and differentiable for all $ x\ne 0 $
D) Continuous and differentiable for all $ x\ne 0 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)=\sqrt[3]{\frac{x^{4}}{| x |}},x\ne 0,f(0)=0 $
$ \therefore f(x)=\sqrt[3]{\frac{x^{4}}{-x}}=\sqrt[3]{-x^{3}}=-xifx<0 $
$ \And f(x)=\sqrt[3]{\frac{x^{4}}{x}}=\sqrt[3]{x^{3}}=xifx>0 $
$ f(x)= \begin{matrix} -x, & if & x<0 \\ 0, & if & x=0 \\ x, & if & x>0 \\ \end{matrix} . $
Clearly f(x) is continuous for all x but not differentiable at x = 0
BETA
