Continuity And Differentiability

Calculus Level 3

The function given by $f(x)=2+\sqrt[3]{x-2}$ at the point $x=2$ is $\text{\_\_\_\_\_\_\_\_\_\_\_\_\_} .$

not continuous nor differentiable differentiable but not continuous continuous but not differentiable continuous and differentiable

1 solution

Tom Engelsman
Feb 26, 2017

The limits $x \rightarrow 2_-$ and $x \rightarrow 2_+$ both exist & approach 2 for $f(x)$ , hence $f(2)$ is continuous. However, $f'(x) = \frac{1}{3(x-2)^{\frac{2}{3}}}$ , and $f'(2)$ is undefined (i.e.not differentiable). Choice D is the correct answer.

How $x \to 2^-$ limit exists.Is it not becoming complex ???

Kushal Bose - 4 years, 3 months ago

Cube root of negative numbers are not complex .

Sabhrant Sachan - 4 years, 3 months ago

Continuity And Differentiability

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