Differentiability & Continuity (College Board AP® Calculus BC): Revision Note

Differentiability & continuity

When is a function differentiable?

• The derivative of a function is defined as $f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

  • The derivative only exists if this limit exists

  • The derivative, if it exists, is itself a function

• If a point is not in the domain of $f\left(x\right)$ then it cannot be in the domain of $f\text{'}\left(x\right)$

  • This means $f\left(x\right)$ will not be differentiable at points where there is a vertical asymptote

    • E.g. $f\left(x\right)=\frac{1}{x}$ is not differentiable at $x=0$

• A differentiable function is one for which its derivative exists at each point in its domain

  • This means that a function with points at which it is not differentiable can still be made a differentiable function by appropriately restricting its domain

How are differentiability and continuity related?

• A function must be continuous at a point to be differentiable at that point

  • Therefore, if a function is differentiable at a point, then it is continuous at that point

  • And if a function is not continuous at a point, then it is not differentiable at that point

• However, if a function is continuous at a point, it is not necessarily differentiable at that point

  • The function could have a corner at a point

    • This is where the one-sided limits of the derivative at the point are finite but unequal

    • E.g. $f\left(x\right)=\left|x\right|$ has a corner at $x=0$

  • The function could have a cusp at a point

    • This is where one of the one-sided limits of the derivative at the point tends to infinity and the other tends to negative infinity

    • E.g. $f\left(x\right)=\sqrt[3]{x^2}$ has a cusp at $x=0$

  • The function could have a vertical tangent at a point

    • This is where both the one-sided limits of the derivative at the point tend to infinity or they both tend to negative infinity

    • E.g. $f\left(x\right)=\sqrt[3]{x}$ has a vertical tangent at $x=0$

  • The function could not be differentiable at a point due to oscillation

    • This is where the derivative oscillates as it approaches the point

    • E.g. $f\left(x\right)=x\sin \left(\frac{1}{x}\right)$ at $x=0$

How can I show a function is not differentiable at a point?

• If a function is not continuous at a point, then it is not differentiable at that point

  • That's the easiest way to show a function is not differentiable at a point!

• If a function is continuous at a point, then to show it is not differentiable

  • you need to go back to the limit definition of the derivative

    • $f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

• Recall that the limit of a function does not exist if the function inside the limit

  • is unbounded at the point in question

  • oscillates near the point in question

  • or has unequal one-sided limits at the point in question

• Consider the graph of $y=\left|x\right|$ shown below

• At (0, 0), $f\left(x\right)=\left|x\right|$ is not differentiable

• To show why, consider the derivative at (0, 0)

  • $f^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0}{\lim }\frac{\left|h\right|-\left|0\right|}{h}=\underset{h\to 0}{\lim }\frac{\left|h\right|}{h}$

• Consider the one-sided limit from the left

  • For $h<0$, $\frac{\left|h\right|}{h}=\frac{-h}{h}=-1$

  • So $\underset{h\to 0^{-}}{\lim }\frac{\left|h\right|}{h}=\underset{h\to 0^{-}}{\lim }\left(-1\right)=-1$

• Consider the one-sided limit from the right

  • For $h>0$, $\frac{\left|h\right|}{h}=\frac{h}{h}=1$

  • So $\underset{h\to 0^{+}}{\lim }\frac{\left|h\right|}{h}=\underset{h\to 0^{+}}{\lim }\left(1\right)=1$

• You can also see this visually from the graph

  • The slope for negative values of $x$ is -1

  • The slope for positive values of $x$ is 1

• The one-sided limits do not agree, therefore the limit $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ at (0, 0) does not exist

  • Therefore $f\left(x\right)=\left|x\right|$ is not differentiable at (0, 0)

• Consider the graph of $y=\sqrt[3]{x}$ shown below

• At (0, 0), $f\left(x\right)=\sqrt[3]{x}$ is not differentiable

• To show why, consider the derivative at (0, 0)

  • $f^{\text{'}}\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{x\to 0}{\lim }\frac{\sqrt[3]{h}}{h}=\underset{h\to 0}{\lim }\frac{1}{\sqrt[3]{h^2}}$

• The limit approaches infinity from both sides, therefore the limit does not exist

  • Therefore $f\left(x\right)=\sqrt[3]{x}$ is not differentiable at (0, 0)