Where a function is not differentiable

What you will be able to do

• find the points where $|g(x)|$, sums of absolute values and products with $|x - a|$ are not differentiable;
• handle the greatest integer function inside products;
• count the points of non-differentiability of a piecewise function by comparing one-sided derivatives.

What to look for

A function can fail to be differentiable in three ways: a corner, a vertical tangent, or a break. Absolute values, greatest integers and joins of pieces create most corners and breaks. The method is always the same.

1. List the candidate points: zeros of absolute values, integers for $[x]$, and joins of pieces.
2. At each candidate, find the left derivative and the right derivative.
3. A point is a point of non-differentiability when they differ, or when one does not exist.

The absolute value of a function

Let $f(x) = |g(x)|$ with $g$ differentiable. Away from the zeros of $g$, $f$ equals $g$ or $-g$ and is differentiable. At a zero $a$ of $g$ there are two cases.

• A simple zero, $g'(a) \ne 0$. The slope of $f$ is $-|g'(a)|$ on one side and $|g'(a)|$ on the other. This is a corner, with a slope jump of $2|g'(a)|$.
• A double zero, $g'(a) = 0$. Now $g(a + h)$ is much smaller than $h$, so the quotient $\dfrac{|g(a + h)|}{h}$ tends to $0$ from both sides. There is no corner.

For example, $|x^2 - 4|$ has corners at $x = \pm 2$, where the slopes jump from $-4$ to $4$ and from $-4$ to $4$ again. But $|(x - 2)^2| = (x - 2)^2$ has no corner at $2$.

A product with $|x - a|$

Let $h(x) = g(x)\,|x - a|$ with $g$ differentiable at $a$. The left derivative at $a$ is $-g(a)$ and the right derivative is $g(a)$. They are equal exactly when $g(a) = 0$, and then both are $0$.

So $(x - 1)|x - 1|$ is differentiable at $1$, but $x|x - 1|$ is not, because $g(1) = 1 \ne 0$. Likewise $|x|^3 = x^2|x|$ is differentiable at $0$, once and twice, but its third derivative has a jump: it is $-6$ for $x < 0$ and $6$ for $x > 0$.

Sums, maxima and minima

• In a sum such as $|x - 1| + |x - 2| + |x - 3|$, each term adds its own slope jump of $2$ at its zero. Since all the jumps are positive, they never cancel. There are corners at $1$, $2$ and $3$.
• For $\max(f, g)$ and $\min(f, g)$, corners appear where the two graphs cross with different slopes. The function $\max(\sin x, \cos x)$ has a corner at $x = \dfrac{\pi}{4}$.
• Take $\sin|x|$. At $0$ the slopes are $-1$ and $1$. It has a corner. But $|\sin x|$ has corners at every multiple of $\pi$.

The greatest integer function inside a product

The function $[x]$ jumps at every integer $n$. In the product $[x]\,g(x)$ the factor is $n - 1$ just to the left of $n$ and $n$ just to the right. The function is continuous at $n$ only if $g(n) = 0$. It is differentiable at $n$ only if the slopes $(n - 1)g'(n)$ and $n\,g'(n)$ agree, which needs $g'(n) = 0$ as well.

So $[x]\sin\pi x$ is continuous at every integer, but never differentiable there, because $g'(n) = \pi\cos\pi n \ne 0$.

A note on words. British and American texts say a graph has a corner or a point where the gradient is not defined. AP also calls a sharp point a cusp when the slopes tend to opposite infinities. NCERT and JEE use "not differentiable at".

A worked example

Find the points where $f(x) = |(x - 1)(x - 2)^2| + |x - 4|$ is not differentiable.

1. The candidates. The zeros inside the absolute values are $1$, $2$ and $4$.
2. At $x = 1$. This is a simple zero of $(x - 1)(x - 2)^2$, since the factor $(x - 2)^2 = 1$ there. So the first term has a corner. The second term is smooth at $1$. The slope jumps from $-2$ to $0$. Not differentiable.
3. At $x = 2$. This is a double zero of $(x - 1)(x - 2)^2$, so the first term is smooth. The second term is smooth at $2$. Differentiable, with slope $-1$.
4. At $x = 4$. The second term has a corner, and the first is smooth. The slope jumps from $15$ to $17$.

The function is not differentiable at $x = 1$ and $x = 4$.

Practise with new numbers

Try it in Python

The program finds the corners of four functions by comparing the left and right slopes at a list of candidate points.

Common mistakes

• Treating every zero of $g$ as a corner of $|g|$. A double zero is smooth.
• Using the product rule at the join. At a point where the formula changes, compare one-sided derivatives.
• Forgetting that $[x]$ can be harmless. In $[x]\,g(x)$ the product is differentiable at $n$ only when both $g(n)$ and $g'(n)$ are zero.
• Missing candidates. A sum of two absolute values has candidates at the zeros of both.

Remember these

• $|g|$ has a corner at each simple zero of $g$ and none at a double zero.
• $g(x)|x - a|$ is differentiable at $a$ exactly when $g(a) = 0$.
• $[x]\,g(x)$ is differentiable at an integer $n$ only if $g(n) = 0$ and $g'(n) = 0$.
• Always compare the left and right derivatives at each candidate point.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. At which points is $|x^2 - 9|$ not differentiable?

2. Is $(x - 2)|x - 2|$ differentiable at $x = 2$?

3. Is $|x|\,(x^2 + 1)$ differentiable at $x = 0$?

4. How many points of non-differentiability does $|\sin x|$ have in $[0, 2\pi]$?

5. Is $x\,[x]$ differentiable at $x = 2$?

References

• OpenStax, Calculus Volume 1, Chapter 3, section on the derivative as a function (where a function is not differentiable).