Is a function differentiable at all points that it is continuous?

Mar 1, 2016

No. Here are 3 examples.

Explanation:

Example 1
$f \left(x\right) = \left\mid x \right\mid$ is continuous but not differentiable at $x = 0$.

(The left and right derivatives are not equal -- there is no tangent line.)

graph{y=absx [-2.75, 2.724, -0.876, 1.862]}

Example 2
$f \left(x\right) = \sqrt[3]{x}$ is continuous but not differentiable at $x = 0$.

($f ' \left(x\right) = \frac{1}{3 \sqrt[3]{{x}^{2}}}$ does not exist at $x = 0$. In fact,)
(${\lim}_{x \rightarrow 0} \left\mid f ' \left(x\right) \right\mid = \infty$ -- the tangent line is vertical.)

graph{root(3)x [-1.596, 1.441, -0.964, 0.555]}

Example 3
$f \left(x\right) = \sqrt[3]{{x}^{2}}$ is continuous but not differentiable at $x = 0$.

($f ' \left(x\right) = \frac{2}{3 \sqrt[3]{x}}$ does not exist at $x = 0$. In fact,)
(${\lim}_{x \rightarrow 0} \left\mid f ' \left(x\right) \right\mid = \infty$ -- the tangent line is vertical.)

graph{x^(2/3) [-1.82, 1.597, -0.343, 1.366]}

I like this third example because it is also an example of a function whose minimum occurs at a critical point at which the derivative does not exist.