Why Radical Functions Need a Domain Check
For a polynomial, every real number is in the domain, and its derivative is defined everywhere. Radical functions require more care. An even root restricts its radicand to nonnegative values, a radical in a denominator cannot be zero, and a derivative formula may fail at a point where the function itself is defined.
These details matter when finding critical numbers. A value that makes a derivative formula zero is only a candidate: it must also belong to the function’s domain. Conversely, a point in the domain can be critical even when the derivative is not zero there, because the derivative may fail to exist.
At an endpoint of a domain, the ordinary two-sided derivative does not exist because the function is not defined on both sides. Under this definition, a domain endpoint is a critical number if it is in the domain. When the derivative formula becomes unbounded near such a point, the difference quotient can confirm that no finite derivative exists.
Differentiate Only Where the Formula Applies
For a positive radicand, the chain rule gives a useful formula. If \(g(x)>0\), then
The condition \(g(x)>0\) is important: the denominator in the formula must be nonzero. The function \(\sqrt{g(x)}\) may still be defined when \(g(x)=0\), but the displayed derivative formula does not decide whether the derivative exists there. That point must be checked separately.
More generally, for a positive integer \(n\), the chain rule gives
where the real-valued expression and its derivative are defined. For even roots, the radicand must be nonnegative for the function to be real-valued; the formula above applies where the radicand is positive. At a zero of the radicand, check the function’s domain and derivative directly rather than assuming the formula applies.
Worked Example: An Interior Critical Number and an Endpoint
Worked Example: An Interior Critical Number and an Endpoint
Find the critical numbers of \(f(x)=\sqrt{x}(x-2)\).
Step 1: Determine the domain. The square root requires \(x\geq 0\), so the domain is \([0,\infty)\).
Step 2: Differentiate on the interior of the domain. For \(x>0\), use the product rule and the derivative of \(\sqrt{x}\):
Since \(2\sqrt{x}>0\) for \(x>0\), this derivative is zero exactly when \(3x-2=0\). Thus \(x=\frac{2}{3}\), which belongs to the domain.
Step 3: Check the domain endpoint \(x=0\). We have \(f(0)=0\). For \(h>0\), the difference quotient at zero is
As \(h\) approaches zero from above, the numerator approaches \(-2\) and the denominator approaches \(0\) through positive values. The quotient tends to \(-\infty\), so there is no finite derivative at \(0\). Since \(0\) is in the domain, it is a critical number.
Check: Substituting \(x=\frac{2}{3}\) into the derivative gives
The critical numbers are therefore \(0\) and \(\frac{2}{3}\). The endpoint is included because it is in the domain and its derivative does not exist.
Why Zeros of the Radicand Need Special Attention
The chain-rule formula for a square root has \(\sqrt{g(x)}\) in its denominator. It is therefore unavailable when \(g(x)=0\). But that fact alone does not prove that the derivative of the original function fails to exist there. The derivative must be checked using its definition when necessary.
For example, consider \(f(x)=\sqrt{x^4}\). Since \(x^4\geq 0\) for every real \(x\), the domain is all real numbers. Also, \(\sqrt{x^4}=x^2\), so \(f'(x)=2x\) everywhere, including \(x=0\). Thus \(0\) is critical because the derivative is zero there, not because the radical fails to be differentiable.
Worked Example: A Zero of the Radicand Where the Derivative Exists
Find the critical numbers of \(f(x)=\sqrt{x^4}\).
Because \(x^4\geq 0\) for all real \(x\), the domain is \(\mathbb{R}\). Simplifying the function gives \(f(x)=x^2\), since \(x^2\geq 0\). Hence \(f'(x)=2x\), and solving \(2x=0\) gives \(x=0\).
Check at the radicand’s zero: The derivative definition at zero gives
As \(h\) approaches zero, this quotient approaches \(0\). Therefore \(f'(0)=0\), agreeing with the derivative \(2x\). The only critical number is \(0\).
Critical Numbers at Domain Endpoints
Radical functions often have domain endpoints at zeros of even-root radicands. If such an endpoint belongs to the domain, check it even though the ordinary derivative formula may be unavailable there. A difference quotient can show whether a finite derivative exists. A derivative that grows without bound as the endpoint is approached is a warning to check, not a substitute for that check.
Worked Example: Two Endpoint Critical Numbers on a Disconnected Domain
Find the critical numbers of \(f(x)=\sqrt{x^2-1}\).
Step 1: Determine the domain. We require \(x^2-1\geq 0\), or \((x-1)(x+1)\geq 0\). Therefore \(x\leq -1\) or \(x\geq 1\), and the domain is \((-\infty,-1]\cup[1,\infty)\).
Step 2: Differentiate where \(x^2-1>0\). On the interior parts of the domain,
This derivative cannot equal zero on those intervals: its numerator \(x\) is nonzero whenever \(x\leq -1\) or \(x\geq 1\).
Step 3: Check the endpoints. Both \(1\) and \(-1\) are in the domain, but the derivative formula is undefined at each. For \(h>0\), the quotient from the right at \(1\) is
which tends to \(+\infty\) as \(h\) approaches zero from above. At \(-1\), approach from the domain on the left by setting \(x=-1-h\), with \(h>0\):
This tends to \(-\infty\). Neither endpoint has a finite derivative. Both belong to the domain, so both are critical numbers. There are no other critical numbers, since the derivative is never zero on the interior of the domain.
Solve First, Then Filter by the Domain
A derivative equation can produce a number that is not a possible input to the original function. Such a number must be discarded. Also, a point excluded from the domain is not critical, even if a derivative expression has a zero or is undefined there.
Worked Example: A Derivative-Zero Candidate Outside the Domain
Find the critical numbers of \(f(x)=\frac{\sqrt{x+1}}{x-2}\).
Step 1: Determine the domain. The square root requires \(x+1\geq 0\), while the denominator requires \(x\ne 2\). Thus the domain is \([-1,\infty)\setminus\{2\}\).
Step 2: Differentiate where \(x>-1\) and \(x\ne 2\). The quotient rule gives
The derivative is zero only when \(-x-4=0\), giving \(x=-4\). But \(-4\) is outside the domain, so it is not a critical number.
Step 3: Check the domain points where the derivative formula fails. The endpoint \(x=-1\) is in the domain and needs checking. Since \(f(-1)=0\), for \(h>0\) its difference quotient is
This tends to \(-\infty\) as \(h\) approaches zero from above, so the derivative does not exist at \(-1\). Therefore \(-1\) is critical. The value \(x=2\) is excluded from the domain, so it cannot be critical. The only critical number is \(-1\).
Check: The only derivative-zero solution was \(-4\), which fails the domain condition. The endpoint check identifies \(-1\), which does satisfy the domain condition.
A Reliable Checklist
For a radical function, use the following process. It keeps domain restrictions and derivative conditions separate, so that neither kind of candidate is overlooked.
Impose the restrictions from even roots and denominators before differentiating.
Use the derivative rules only at points where the original function and the derivative formula are defined.
Solve \(f'(x)=0\), then discard solutions outside the original domain.
Check endpoints and any in-domain points where the derivative formula is undefined. Use the derivative definition if needed.
A point is critical if it belongs to the domain and the derivative is zero or does not exist there.
Common Mistake: Treating Every Denominator Zero as Critical
In the last example, the derivative formula has a denominator involving \(x-2\), but \(x=2\) is not a critical number. The original function is not defined there. The definition requires a critical number to be in the domain. In contrast, \(x=-1\) is in the domain, and its derivative does not exist, so it is critical.
Finding critical numbers is a necessary step in studying the behavior of a function, but it does not by itself classify a point as a maximum or minimum. That requires additional analysis. For radical functions, the essential work here is to combine derivative calculations with an exact account of where the function exists.
Check Your Understanding
For each question, state the relevant domain restriction as well as your conclusion about critical numbers.
- For \(f(x)=\sqrt{x}(x+1)\), find the domain and determine whether \(x=0\) is critical.
- For \(f(x)=\sqrt{x^2-4}\), identify the domain endpoints and explain why each must be checked.
- Why is a solution to \(f'(x)=0\) not automatically a critical number?
- Can a function be differentiable at a point where the radicand of a radical is zero? Explain using \(f(x)=\sqrt{x^4}\).
- For \(f(x)=\frac{\sqrt{x}}{x-1}\), explain why \(x=1\) cannot be a critical number.
