Why Absolute Values Suggest Breakpoints
An absolute-value expression can change its formula when the expression inside the bars changes sign. For example, \(|x-2|\) equals \(2-x\) to the left of \(2\) and \(x-2\) to the right. The two formulas meet at \(x=2\), but their slopes differ, so the graph has a corner there.
For a more complicated function, each zero inside an absolute value is a candidate breakpoint. The word candidate matters: a zero may produce a corner, but it need not. To decide, locate the zeros, determine the formulas on the intervals around them, and compare the one-sided slopes of the full function.
When Can the Absolute-Value Bars Be Removed?
The basic identity is \(|u|=u\) when \(u\ge 0\), and \(|u|=-u\) when \(u<0\). To use it on an interval, we need to know the sign of the inside expression throughout that interval.
Continuity gives a useful guarantee. If \(g\) is continuous at \(a\) and \(g(a)>0\), then \(g(x)>0\) for all \(x\) sufficiently close to \(a\). If \(g(a)<0\), then \(g(x)<0\) sufficiently close to \(a\). Thus, if \(g\) is continuous and nonzero at \(a\), its sign is fixed in some neighborhood of \(a\). Likewise, a continuous function with no zeros on an interval has a fixed sign there.
This condition cannot be omitted: without continuity, a function may be nonzero at every point yet have a sign that varies arbitrarily close to a point. In that case, the absolute-value bars cannot necessarily be removed using one formula on a neighborhood.
On each such interval, replace every absolute value by the appropriate signed expression. Then test each candidate breakpoint using the two one-sided derivatives. A function is differentiable at an interior point \(a\) only if both one-sided derivatives exist as finite numbers and are equal.
A General Slope Test at a Zero
The derivative of \(g\) at \(a\), when it exists, is the limit of \(\frac{g(a+h)-g(a)}{h}\) as \(h\) approaches zero. Suppose \(g(a)=0\) and \(g\) is differentiable at \(a\). For the absolute value \(|g(x)|\), the right-hand difference quotient is \(\frac{|g(a+h)|}{h}\) for \(h>0\). The left-hand difference quotient has \(h<0\), so its sign is reversed when the denominator is written as \(-h>0\).
The equalities use \(g(a)=0\) and the fact that the absolute-value function is continuous: if a quantity approaches \(g'(a)\), its absolute value approaches \(|g'(a)|\). Thus a single term \(|g(x)|\) has opposite one-sided slopes at a zero whenever \(g'(a)\ne 0\). If \(g'(a)=0\), both one-sided slopes are zero; that term is differentiable there, provided the stated derivative exists.
Other terms in the function can affect the slopes of the total. For instance, if \(q\) is differentiable at \(a\), then \(q(x)+|g(x)|\) has one-sided slopes \(q'(a)-|g'(a)|\) and \(q'(a)+|g'(a)|\), respectively, under these assumptions. In a sum with several absolute values, compare the one-sided slopes after accounting for every term.
Worked Examples
Worked Example: A Single Absolute Value of a Linear Expression
Find the breakpoint and test differentiability for \(f(x)=|2x-4|\).
Set the inside expression equal to zero: \(2x-4=0\), so \(x=2\). For \(x<2\), \(2x-4<0\), and for \(x>2\), \(2x-4>0\). Therefore,
The left formula has slope \(-2\), and the right formula has slope \(2\). To check these directly at the breakpoint, first note \(f(2)=0\). For \(h<0\), \(f(2+h)=|2h|=-2h\), so the left-hand difference quotient is \(\frac{-2h}{h}=-2\). For \(h>0\), \(f(2+h)=|2h|=2h\), so the right-hand difference quotient is \(\frac{2h}{h}=2\).
The one-sided slopes differ, so \(f\) is not differentiable at \(x=2\). The function is continuous there because both formulas give the value \(0\) at \(2\); continuity alone does not make the slopes match.
Worked Example: Several Breakpoints in a Sum
Find all breakpoint candidates and test differentiability for \(F(x)=|x+2|+|x-1|\).
The inside expressions vanish at \(x=-2\) and \(x=1\). These points divide the real line into three intervals. If \(x<-2\), both expressions are negative. If \(-2<x<1\), \(x+2\) is positive and \(x-1\) is negative. If \(x>1\), both are positive. Consequently,
At \(x=-2\), the left-hand slope is \(-2\) and the right-hand slope is \(0\). The formulas also meet: the left formula gives \(-2(-2)-1=3\), and the middle formula gives \(3\). The function is continuous at \(-2\), but the unequal slopes show it is not differentiable there.
At \(x=1\), the left-hand slope is \(0\) and the right-hand slope is \(2\). Again the formulas meet: the middle value is \(3\), and the right formula gives \(2(1)+1=3\). The unequal slopes show that \(F\) is not differentiable at \(1\) either.
Thus both zeros are actual nondifferentiability points. The interval formulas provide the slopes, and checking their shared endpoint values confirms that these joins are continuous.
Worked Example: A Zero That Does Not Create a Corner
Test differentiability at \(x=0\) for \(H(x)=|x^2|\).
The inside expression \(x^2\) is zero only at \(0\). For every \(x\), \(x^2\ge 0\), so \(|x^2|=x^2\) on both sides of zero. Also \(H(0)=0\). Its difference quotient at zero is
As \(h\) approaches zero from either side, \(h\) approaches \(0\). Both one-sided derivatives therefore equal \(0\), so \(H'(0)=0\). This verifies that a zero inside an absolute value is only a candidate breakpoint: here the inside expression does not change sign, and the function is differentiable at the zero.
As a check, the resulting formula is \(H(x)=x^2\), and the difference quotient just computed tends to the claimed derivative \(0\) at \(x=0\).
A Reliable Breakpoint Procedure
For a function built from absolute-value expressions, use this order of steps. At each candidate, analyze the whole function, not just one absolute-value term.
Set the inside of each absolute value equal to zero and solve within the function’s domain.
Use test values or sign reasoning. When relying on a sign staying fixed nearby, ensure the inside expression is continuous and nonzero there.
Remove each absolute-value bar with the sign appropriate to that interval.
At each interior candidate, calculate the left- and right-hand derivatives of the full function. The derivative exists only if both are finite and equal.
This procedure is useful because it reduces a piecewise-looking expression to familiar formulas on intervals. It also keeps two separate questions clear: whether the function values meet at the breakpoint, and whether the slopes match there.
Common Mistake: Using the Variable’s Sign Instead
A frequent error is to decide the sign of an inside expression from the sign of \(x\), rather than from the expression itself. For \(x<0\), someone might incorrectly claim \(x^2<0\), then write \(|x^2|=-x^2\) on the left of zero and \(x^2\) on the right. That reasoning confuses the sign of \(x\) with the sign of \(x^2\).
The correction is to test the inside expression: \(x^2\ge 0\) on both sides of zero, and \(x^2>0\) whenever \(x\ne0\). Therefore \(|x^2|=x^2\) on both sides, not a pair of different formulas. Its difference quotient at zero is \(h\), which tends to \(0\) from both sides.
Check Your Understanding
For each question, identify the relevant breakpoint candidates and use interval formulas or one-sided difference quotients to justify your conclusion.
- For \(f(x)=|x-3|\), what are the one-sided slopes at its breakpoint? Is the function differentiable there?
- For \(g(x)=|x+1|+|x-2|\), find the breakpoint candidates and write a formula on each interval.
- Why does continuity of an inside expression matter when removing absolute-value bars near a point where it is nonzero?
- For \(h(x)=|x^2-4|\), locate the candidates and determine the signs of the inside expression on the intervals they create.
- Does a zero inside an absolute value always create a corner? Give a reason based on an example from this tutorial.
