Why Absolute Values Need Special Attention
An absolute value can change the derivative abruptly where its inside expression changes sign. For instance, \(f(x)=|x|\) has a sharp corner at \(x=0\): its slope approaching from the left differs from its slope approaching from the right. That point matters when identifying critical numbers, even though the derivative there is not zero.
Not every zero inside an absolute value creates a corner. If the inside expression touches zero without crossing it, the absolute-value function may still be differentiable there. A reliable analysis therefore asks three questions: Is the candidate in the domain? Do the one-sided derivatives agree? And what does the function’s value imply about a local extremum?
Critical Numbers and Domain Membership
Here, \(f'(c)\) means the ordinary two-sided derivative. At an interior point, a mismatch between finite left and right derivatives means that the derivative does not exist. A domain endpoint can also be a critical number under this definition if it belongs to the domain and the ordinary derivative does not exist there.
For an absolute-value function \(f(x)=|g(x)|\), a useful first step is to find where \(g(x)=0\). These zeros are potential corners, but they must be checked: a zero might be outside the domain, or the graph might pass smoothly through it.
Derivatives Away from Zeros
At points where \(g(x)\ne 0\), the sign of \(g(x)\) stays fixed in a sufficiently small neighborhood. If \(g(x)>0\), then \(|g(x)|=g(x)\); if \(g(x)<0\), then \(|g(x)|=-g(x)\). The chain rule therefore gives the following formula wherever \(g\) is differentiable and nonzero:
Equivalently, at such points the derivative is \(\operatorname{sgn}(g(x))g'(x)\), where \(\operatorname{sgn}(u)=1\) for \(u>0\) and \(\operatorname{sgn}(u)=-1\) for \(u<0\). This formula does not apply at a zero of \(g\); there we need to examine the derivative directly.
What Happens at a Zero
Suppose \(g\) is differentiable at \(c\) and \(g(c)=0\). The left and right derivatives of \(|g|\) at \(c\), when they are considered as limits, are governed by the slope \(g'(c)\). The result below explains exactly when a corner occurs.
To justify the result, write the difference quotient for \(h\ne 0\). Since \(g(c)=0\), it is:
Differentiability of \(g\) at \(c\) means \(g(c+h)=g'(c)h+o(h)\), where \(o(h)/h\) tends to zero as \(h\) tends to zero. For \(h>0\), taking the limit of the quotient gives \(|g'(c)|\). For \(h<0\), division by a negative number reverses the sign, so the limit is \(-|g'(c)|\). If \(g'(c)\ne 0\), these values differ, and the two-sided derivative does not exist. If \(g'(c)=0\), both values are zero, so the two-sided derivative exists and equals zero. This proves the claim.
At any zero \(c\) that belongs to the domain, \(|g(c)|=0\), while \(|g(x)|\ge 0\) everywhere in the domain. Consequently, \(c\) is a global minimum of \(|g|\) on its domain, whether or not the graph has a corner there.
Worked Examples
Worked Example: A Simple Corner
Find and classify the critical numbers of \(f(x)=|3x-6|\).
The function is defined for every real number. Its inside expression is zero when \(3x-6=0\), so \(x=2\) is the candidate corner. Set \(g(x)=3x-6\); then \(g'(x)=3\), so \(g'(2)=3\ne 0\). The theorem gives left derivative \(-|3|=-3\) and right derivative \(|3|=3\). The derivatives differ, so \(f'(2)\) does not exist and \(2\) is a critical number.
Also, \(f(2)=|0|=0\), and \(f(x)\ge 0\) for every real \(x\). Therefore \(x=2\) is a global minimum. To check the slopes directly, for \(h<0\), \(f(2+h)=|3h|=-3h\), so \(\bigl(f(2+h)-f(2)\bigr)/h=-3\). For \(h>0\), \(f(2+h)=3h\), so the quotient is \(3\). This confirms the one-sided derivatives.
Worked Example: Corners and a Stationary Point
Find all critical numbers of \(f(x)=|x^2-1|\), and identify the local extrema among them.
The domain is all real numbers. Let \(g(x)=x^2-1\), so its zeros are \(x=-1\) and \(x=1\), and \(g'(x)=2x\). At \(x=-1\), \(g'(-1)=-2\), so the left derivative of \(f\) is \(-|-2|=-2\) and the right derivative is \(|-2|=2\). At \(x=1\), \(g'(1)=2\), so the left derivative is \(-|2|=-2\) and the right derivative is \(|2|=2\). Both points are corners and therefore critical numbers.
For \(x\ne -1,1\), the derivative formula away from zeros applies. When \(x^2-1>0\), \(f'(x)=2x\); when \(x^2-1<0\), \(f'(x)=-2x\). In the second region, which is \(-1<x<1\), the derivative is zero at \(x=0\). Since \(f(0)=|{-1}|=1\), \(0\) is another critical number. There are no other zeros of the derivative away from \(x=-1,1\), so the critical numbers are \(-1,0,1\).
At \(x=-1\) and \(x=1\), the function equals zero and is nonnegative, so both points are global minima. Near zero, \(f(x)=1-x^2\) for \(-1<x<1\). For nonzero \(x\) in this interval, \(1-x^2<1=f(0)\), so \(0\) is a strict local maximum. As a check, differentiating \(1-x^2\) gives \(-2x\), which is zero at \(0\), as found above.
Worked Example: Domain Endpoints and a Corner
Find the critical numbers of \(f(x)=|\sqrt{x+1}-2|\), and classify its extrema on its domain.
The square root requires \(x+1\ge 0\), so the domain is \([-1,\infty)\). Set \(g(x)=\sqrt{x+1}-2\). Its zero satisfies \(\sqrt{x+1}=2\), hence \(x+1=4\) and \(x=3\). At this interior point, \(g'(x)=1/(2\sqrt{x+1})\), so \(g'(3)=1/4\). The left derivative of \(f\) at \(3\) is \(-1/4\), and the right derivative is \(1/4\). Thus \(3\) is a critical number and a global minimum, since \(f(3)=0\) and \(f(x)\ge 0\).
The domain endpoint \(x=-1\) must also be checked. It belongs to the domain, and \(f(-1)=|0-2|=2\). For \(h>0\), \(f(-1+h)=|\sqrt h-2|\). When \(h\) is sufficiently small, \(\sqrt h<2\), so \(f(-1+h)=2-\sqrt h\). The right difference quotient is then \(\bigl(f(-1+h)-f(-1)\bigr)/h=-1/\sqrt h\), which has no finite limit as \(h\) tends to zero from the right. The ordinary derivative at the endpoint does not exist, so \(-1\) is also a critical number. Since \(2-\sqrt h<2\) for \(h>0\), values immediately to the right are smaller than \(f(-1)\); thus \(-1\) is a relative-domain local maximum. The critical numbers are \(-1\) and \(3\).
A Domain Check Can Remove a Candidate
A zero of the expression inside an absolute value is only a candidate if the whole function is defined there. For example, consider \(q(x)=\left|\frac{x^2-1}{x-1}\right|\). Its domain excludes \(x=1\), even though \(x^2-1=0\) there. Thus \(x=1\) is not a critical number: the function has no value at that point. At \(x=-1\), however, the function is defined and its inside expression is zero. The quotient inside has derivative \(1\) at \(-1\), so the absolute value has left derivative \(-1\) and right derivative \(1\) there. Hence \(-1\) is a critical number and a global minimum.
Common Mistake: Differentiating Through a Corner
A tempting but incorrect calculation for \(f(x)=|x^2-1|\) is \(f'(x)=2x\) everywhere. This differentiates the inside expression but ignores the absolute value. At \(x=0\), for example, the correct derivative is \(-2x\) in the interval \((-1,1)\), so \(f'(0)=0\), not \(2(0)\) by a rule that was applied everywhere. At \(x=1\), neither formula for nonzero inside values applies: the one-sided derivatives are \(-2\) and \(2\), so \(f'(1)\) does not exist.
A Reliable Analysis
Record restrictions from denominators, radicals, and any other parts of the formula.
Keep only zeros that are in the domain as possible corners or smooth critical points.
If the inside function is differentiable there, compare the one-sided derivatives. A nonzero inside derivative gives unequal slopes; a zero inside derivative gives derivative zero.
Away from zeros, use the derivative formula for the fixed sign of the inside expression. Include domain points where the ordinary derivative does not exist.
Every domain point where the absolute-value function is zero is a global minimum, since absolute values are nonnegative.
Why These Checks Matter
Absolute-value graphs often have visually prominent corners, and a corner can be a critical number even though its derivative is not zero. But visual sharpness alone is not a definition: the point must belong to the domain, and its one-sided derivatives determine whether a derivative exists. Moreover, zeros need not be corners. When the inside function has derivative zero at a zero, the absolute-value function is differentiable there, as the theorem shows.
Keeping domain, differentiability, and extremum classification separate prevents several errors. A point can be excluded from the domain, a critical number can be a smooth stationary point rather than a corner, and a zero of an absolute-value function is a minimum whether the graph is sharp or smooth.
Check Your Understanding
For each question, justify your conclusion using the domain, derivative, or one-sided derivative as appropriate.
- For \(f(x)=|5x+10|\), find the corner and its left and right derivatives.
- For \(f(x)=|x^2|\), is the function differentiable at its zero? Is that point a critical number?
- For \(f(x)=|x^2-4|\), identify its critical numbers and state which are zeros of the function.
- Why is \(x=1\) not a critical number of \(q(x)=\left|\frac{x^2-1}{x-1}\right|\)?
- Can a domain endpoint be a critical number under the ordinary two-sided derivative definition? Explain the role of domain membership.
