Why Power Functions Have Distinct Concavity Patterns
Power functions such as \(x^3\), \(x^{2/3}\), and \(x^{-1/3}\) can look similar in notation but have different domains and shapes. Their concavity is governed by the exponent, while their behavior at zero also depends on whether the function is defined there and on both sides of it.
The key calculation is the second derivative. The key interpretation is a sign change: an inflection point requires the graph to change concavity, not merely for the second derivative to be zero or undefined. This distinction is especially important for fractional powers, where the second derivative may fail to exist at zero.
Power Functions and Their Domains
For \(x>0\), the power function \(f(x)=x^p\) is defined for any real exponent \(p\). For negative inputs, fractional powers require more care. Write a rational exponent in lowest terms as \(p=m/n\), where \(m\) is an integer and \(n\) is a positive integer. If \(n\) is odd, \(x^{m/n}\) is real for negative \(x\), using the real \(n\)th root. If \(n\) is even, it is not real for negative \(x\).
At \(x=0\), positive powers are defined by \(0^p=0\) when \(p>0\). Negative powers are not defined there. The zero exponent gives the constant function \(1\) on its usual domain. These domain differences determine whether concavity on the two sides of zero can be compared.
The Second Derivative Sign Test
On the positive half-line, the power rule applies twice. For \(f(x)=x^p\),
Since \(x^{p-2}>0\) whenever \(x>0\), the sign of \(f''\) there is the sign of \(p(p-1)\). This gives the basic classification below. The cases \(p=0\) and \(p=1\) have zero second derivative and are constant or linear, respectively.
| Exponent \(p\) | Concavity for \(x>0\) |
|---|---|
| \(p<0\) | Concave up, since \(p(p-1)>0\) |
| \(0<p<1\) | Concave down, since \(p(p-1)<0\) |
| \(p>1\) | Concave up, since \(p(p-1)>0\) |
| \(p=0\) or \(p=1\) | Constant or linear; no change in concavity |
When \(p=m/n\) is in lowest terms with \(n\) odd, the same derivative formula holds for \(x<0\), wherever the power function is defined and differentiable. On that half-line, \(x^{p-2}\) has the sign of \((-1)^m\): subtracting \(2\) from \(m/n\) changes the numerator to \(m-2n\), which has the same parity as \(m\). Thus, for \(p\ne0,1\), the sign of \(f''\) on the negative half-line is the positive-side sign multiplied by \((-1)^m\).
Why the Sign Rule Works
The rule follows directly from the second derivative, together with the sign of a rational power. The derivation also shows why parity matters on negative inputs.
For nonzero \(x\) in the real domain, the power rule gives \(f''(x)=p(p-1)x^{p-2}\).
If \(x>0\), then \(x^{p-2}>0\). Therefore \(f''\) has the sign of \(p(p-1)\).
For \(p=m/n\) in lowest terms with \(n\) odd, \(x^{p-2}=x^{(m-2n)/n}\) is negative exactly when \(m-2n\) is odd. Because \(2n\) is even, this is exactly when \(m\) is odd.
This is a sign analysis on each part of the domain. It does not by itself establish an inflection point at zero: zero must belong to the domain, and the signs on the two sides must actually differ.
Worked Example: An Integer Power with an Inflection Point
Worked Example: An Integer Power with an Inflection Point
Classify the concavity and inflection behavior of \(f(x)=x^3\).
The domain is all real numbers. Differentiating gives
If \(x<0\), then \(6x<0\), so the function is concave down. If \(x>0\), then \(6x>0\), so it is concave up. The concavity changes at zero, and \(f(0)=0\). Therefore \((0,0)\) is an inflection point.
Check the derivatives: differentiating \(x^3\) gives \(3x^2\), and differentiating \(3x^2\) gives \(6x\). Substitution gives \(f(0)=0^3=0\), confirming the point on the graph. Here \(f''(0)=0\), but the inflection conclusion comes from the sign change, not from the zero alone.
What Happens at Zero?
Suppose \(p>0\) and \(p=m/n\) is in lowest terms with \(n\) odd. The function is defined on both sides of zero and is continuous at zero, with \(f(0)=0\). For \(p\ne1\), the concavity signs on the two sides differ exactly when \(m\) is odd, because the negative-side sign is the positive-side sign multiplied by \((-1)^m\). Thus, in this setting, zero is an inflection point when \(m\) is odd; when \(m\) is even, the concavity is the same on both sides. The linear case \(p=1\) is an exception to the sign test because \(p(p-1)=0\), and the function has no change in concavity.
If \(n\) is even and \(p>0\), the real domain begins at zero and extends to the right. Zero is then an endpoint, not an interior point with graph on both sides. The function can be concave up or down to the right, but zero is not an inflection point under the definition above.
For \(p<0\), zero is not in the domain. Even if the concavity on the negative and positive half-lines differs, there is no point on the graph at zero where the function changes concavity. A gap or vertical asymptote cannot serve as an inflection point.
Worked Example: A Fractional Power with an Undefined Second Derivative
Classify \(f(x)=x^{5/3}\), including its behavior at zero.
The denominator \(3\) is odd, so the domain includes negative inputs as well as zero. For \(x\ne0\),
For \(x<0\), \(x^{-1/3}<0\), so \(f''(x)<0\). For \(x>0\), \(x^{-1/3}>0\), so \(f''(x)>0\). The function changes from concave down to concave up at zero.
The function is continuous at zero, with \(f(0)=0\). Its first derivative there exists: the difference quotient is \(\frac{f(h)-f(0)}{h}=\frac{h^{5/3}}{h}=h^{2/3}\), which tends to \(0\) as \(h\) tends to \(0\). However, the displayed formula for \(f''\) is undefined at zero. Since the concavity changes and the graph is continuous there, \((0,0)\) is an inflection point despite the undefined second derivative.
Check the differentiation for \(x\ne0\): differentiating \(\frac{5}{3}x^{2/3}\) gives \(\frac{5}{3}\cdot\frac{2}{3}x^{-1/3}=\frac{10}{9}x^{-1/3}\), as used in the sign test.
Worked Example: No Inflection Despite an Undefined Second Derivative
Classify \(f(x)=x^{2/3}\) at and around zero.
The denominator is odd, so the function is real for all \(x\), and \(f(0)=0\). For \(x\ne0\), differentiation gives
For every nonzero \(x\), \(x^{-4/3}\) is positive, so \(f''(x)<0\) on both sides of zero. The graph is concave down on each side and does not change concavity. Zero is not an inflection point, even though \(f''\) is undefined there.
In fact, \(f'\) is also undefined at zero: its difference quotient is \(h^{2/3}/h=h^{-1/3}\), which has no finite two-sided limit. This does not affect the concavity conclusion, which follows from the second-derivative signs on either side. Checking the differentiation, \(\frac{2}{3}\cdot(-\frac{1}{3})=-\frac{2}{9}\), confirming the coefficient.
When Negative Powers Have a Gap at Zero
Worked Example: Different Concavity on Disconnected Intervals
Classify \(f(x)=x^{-1/3}\) and decide whether zero is an inflection point.
The domain is all nonzero real numbers. For \(x\ne0\),
For \(x<0\), \(x^{-7/3}<0\), so \(f''<0\) and the function is concave down. For \(x>0\), \(x^{-7/3}>0\), so \(f''>0\) and the function is concave up. But zero is not in the domain, so there is no point on the graph there. The two concavity intervals are separated by a vertical asymptote, not joined at an inflection point.
Check the derivatives: the coefficient in the second derivative is \((-\frac{1}{3})(-\frac{4}{3})=\frac{4}{9}\), and differentiating the power lowers its exponent from \(-4/3\) to \(-7/3\). The sign conclusions follow from the odd denominator and odd numerator of that exponent.
A Common Mistake: Treating Every Candidate as an Inflection
A frequent error is to label a point an inflection point merely because \(f''\) is zero or undefined there. For example, the calculation for \(x^{2/3}\) gives an undefined second derivative at zero. Calling zero an inflection point on that basis is incorrect: \(f''<0\) on both sides, so the concavity does not change.
The reverse error is to rule out an inflection point because \(f''\) is undefined. The example \(x^{5/3}\) shows why this fails: its second derivative is undefined at zero, yet the signs on either side differ and the function is continuous there. In both cases, the correction is to test concavity on the intervals to the left and right, then check that the candidate point belongs to the graph.
Check Your Understanding
For each question, use the domain as well as the second-derivative signs in your reasoning.
- On what intervals is \(x^{7/3}\) concave up or concave down, and is zero an inflection point?
- Why is zero not an inflection point of \(x^{4/3}\), even though the second derivative is undefined there?
- For \(x^{-2}\), classify concavity on each part of the domain. Can zero be an inflection point?
- For a positive exponent \(p=m/n\) in lowest terms with \(n\) odd, what role does the parity of \(m\) play in whether concavity changes at zero?
- What additional check is required before calling a point where \(f''\) is undefined an inflection point?
