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 line to the graph \( G \). We can say that \( f \) is not differentiable for any value of \( x \) where a tangent cannot exist, or where a tangent exists but is vertical (a vertical line has an undefined slope, hence an undefined derivative).
Below are graphs of functions that are not differentiable at \( x = 0 \) for various reasons:
Example 1: Sharp Corner (No Tangent)
Function \( f \) below is not differentiable at \( x = 0 \) because there is no unique tangent to the graph at \( x = 0 \) (try to draw a tangent at \( x = 0 \)!)
Example 2: Cusp / Sharp Point
Function \( g \) below is not differentiable at \( x = 0 \) because there is no tangent to the graph at \( x = 0 \).
Example 3: Jump Discontinuity
Function \( h \) below is not differentiable at \( x = 0 \) because there is a jump in the value of the function, and the function is not continuous at \( x = 0 \).
Example 4: Vertical Asymptote
Function \( j \) below is not differentiable at \( x = 0 \) because it increases indefinitely (no limit) on each side of \( x = 0 \), and from its formula is undefined and thus discontinuous at \( x = 0 \).
Example 5: Vertical Tangent
Function \( k \) below is not differentiable because the tangent at \( x = 0 \) is vertical, and therefore its slope (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 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 \( x = a \), it does not necessarily mean \( f \) 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: Piecewise Function Non-Differentiability
Problem: 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:
One way to answer the question is to calculate the derivative at \( x = 0 \) using the limit of the difference quotient. Since function \( f \) is defined using different formulas, we need to evaluate the derivative using left-hand and right-hand limits:
\[ f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} \]On the left of \( x = 0 \) (\( x < 0 \)), the derivative is calculated as follows:
\[ f'(0) = \lim_{h\to 0^{\mathbf{-}}} \frac{f(0+h) - f(0)}{h} = \lim_{h\to 0^-} \frac{-h - 0}{h} = -1 \]On the right of \( x = 0 \) (\( x > 0 \)), the derivative is calculated as follows:
\[ f'(0) = \lim_{h\to 0^{\mathbf{+}}} \frac{f(0+h) - f(0)}{h} = \lim_{h\to 0^+} \frac{h^2 - 0}{h} = \lim_{h\to 0^+} h = 0 \]The limits to the left and to the right of \( x = 0 \) are not equal (\(-1 \neq 0\)), and therefore \( f'(0) \) is undefined. Hence, function \( f \) is not differentiable at \( x = 0 \).
The graph of function \( f \) is shown below, where it is clear that no unique tangent can be drawn at \( x = 0 \):