# Non Differentiable Functions

Questions on the differentiability of functions with emphasis on piecewise functions are presented along with their answers.

## Graphical Meaning of non differentiability.

Which Functions are non Differentiable?
Let f be a function whose graph is G. From the definition, the value of the derivative of a function f at a certain value of x is equal to the slope of the tangent to the graph G. We can say that f is not differentiable for any value of x where a tangent cannot 'exist' or the tangent exists but is vertical (vertical line has undefined slope, hence undefined derivative).
Below are graphs of functions that are not differentiable at $$x = 0$$ for various reasons.
Function f below is not differentiable at $$x = 0$$ because there is no tangent to the graph at $$x = 0$$.(try to draw a tangent at x=0!)
Function g below is not differentiable at $$x = 0$$ because there is no tangent to the graph at $$x = 0$$.(try to draw a tangent at x=0!)
Function h below is not differentiable at $$x = 0$$ because there is a jump in the value of the function and also the function is not defined therefore not continuous at $$x = 0$$.
Function j below is not differentiable at $$x = 0$$ because it increases indefinitely (no limit) on each sides of $$x = 0$$ and also from its formula is undefined at $$x = 0$$ and therefore non continuous at $$x=0$$ .
Function k below is not differentiable because the tangent at $$x = 0$$ is vertical and therefore its slope which the value of the derivative at $$x =0$$ is undefined.

## Theorem

Theorem: If a function f is differentiable at $$x = a$$, then it is continuous at $$x = a$$
Contrapositive of the above theorem: If function f is not continuous at $$x = a$$, then it is not differentiable at $$x = a$$.
Common mistakes to avoid: If f is continuous at $$x = a$$, then f is differentiable at $$x = a$$.
NOTE: Although functions f, g and k (whose graphs are shown above) are continuous everywhere, they are not differentiable at $$x = 0$$.

## Examples with Solutions

Analytical Proofs of non differentiability

Example 1: Show analytically that function f defined below is non differentiable at $$x = 0$$.
$$f(x) = \begin{cases} x^2 & x > 0 \\ - x & x < 0 \\ 0 & x = 0 \end{cases}$$

Solution to Example 1
One way to answer the above question, is to calculate the derivative at $$x = 0$$. We start by finding the limit of the difference quotient. Since function f is defined using different formulas, we need to find the derivative at $$x = 0$$ using the left and the right limits.
$$f'(x) = \lim_{h\to 0} \dfrac{f(x+h) - f(x)}{h}$$
On the left of $$x = 0$$ ($$x \lt 0$$), the derivative is calculated as follows
$$f'(0) = \lim_{h\to 0^-} \dfrac{f(0+h) - f(0)}{h} = \lim_{h\to 0} \dfrac{ -h - 0}{h} = -1$$
On the right of $$x = 0$$ ($$x > 0$$), the derivative is calculated as follows
$$f'(0) = \lim_{h\to 0^+} \dfrac{f(0+h) - f(0)}{h} = \lim_{h\to 0} \dfrac{h^2 - 0}{h} = \lim_{h\to 0} h = 0$$
The limits to the left and to the right of $$x = 0$$ are not equal and therefore $$f '(0)$$ is undefined and hence function $$f$$ in not differentiable at $$x = 0$$.
The graph of function $$f$$ solved in this example is shown below and it is easy to note that no tangent can be drawn at $$x = 0$$ and hence $$f$$ is not differentiable at x = 0.