Why a Zero Derivative Is Not Enough
A derivative of zero tells us that the graph has a horizontal tangent at that point, provided the derivative exists. It does not, by itself, tell us whether the function turns there.
A function may be increasing before and after a stationary point. In that case, the graph briefly has a horizontal tangent but continues in the same direction. The point is neither a local maximum nor a local minimum.
To decide what happens, examine the derivative on both sides of the stationary point. The derivative’s value at the point is only one piece of information; its nearby signs reveal whether the function changes direction.
Recall that \(c\) is a local maximum number if \(f(c)\geq f(x)\) for every \(x\) in some open interval around \(c\). It is a local minimum number if \(f(c)\leq f(x)\) for every \(x\) in some open interval around \(c\). A local extremum means either a local maximum or a local minimum.
Classify the Point by Testing Both Sides
Suppose \(f'(c)=0\). Choose a small interval around \(c\) and determine the sign of \(f'(x)\) for \(x<c\) and \(x>c\). A derivative that is positive means the function is increasing on the interval in question; a negative derivative means it is decreasing.
| Sign to the left of \(c\) | Sign to the right of \(c\) | What the function does near \(c\) |
|---|---|---|
| Positive | Positive | Continues increasing; no local extremum if the signs are strictly positive on both sides. |
| Negative | Negative | Continues decreasing; no local extremum if the signs are strictly negative on both sides. |
| Positive | Negative | Changes from increasing to decreasing; \(c\) is a local maximum. |
| Negative | Positive | Changes from decreasing to increasing; \(c\) is a local minimum. |
The important feature in this tutorial is the first two rows: the derivative has the same strict sign on each side. The function keeps moving in one direction through the stationary point instead of turning.
Why Unchanged Derivative Signs Rule Out an Extremum
The reasoning follows from the Mean Value Theorem. If \(f\) is continuous on a closed interval and differentiable in its interior, then the change in \(f\) between the endpoints equals the interval length multiplied by a derivative value at some interior point.
Suppose \(f'(c)=0\), and suppose \(f'(x)>0\) for all \(x\) sufficiently close to \(c\), except possibly at \(c\). For any nearby \(x>c\), the Mean Value Theorem applied on \([c,x]\) gives a number \(\xi\in(c,x)\) such that
Here \(f'(\xi)>0\) and \(x-c>0\), so \(f(x)-f(c)>0\). Thus \(f(x)>f(c)\) to the right. For a nearby \(x<c\), apply the theorem on \([x,c]\). There is a \(\xi\in(x,c)\) such that
Both factors on the right are positive, so \(f(c)>f(x)\) to the left. Values are below \(f(c)\) on the left and above \(f(c)\) on the right. Therefore \(c\) is neither a local maximum nor a local minimum.
If instead \(f'(x)<0\) on both punctured sides, the same argument gives \(f(x)<f(c)\) to the right and \(f(x)>f(c)\) to the left. Again, \(c\) is neither type of local extremum.
This argument uses strict signs in a neighborhood on each side. Merely knowing that the derivative is nonnegative or nonpositive is not enough to make the same conclusion in every situation: the derivative could be zero on part of an interval. For the examples below, the derivative has a strict, unchanged sign at all nearby points other than the stationary number.
Worked Example: A Basic Cubic Stationary Point
Worked Example: A Basic Cubic Stationary Point
Classify the stationary number of \(f(x)=x^3\).
Step 1: Find the derivative and stationary number. Differentiating gives
Set the derivative equal to zero:
Thus \(c=0\) is the stationary number, and \(f(0)=0^3=0\). The stationary point is \((0,0)\).
Step 2: Test the derivative signs on both sides. If \(x<0\), then \(x^2>0\), so \(3x^2>0\). If \(x>0\), then \(x^2>0\), so \(3x^2>0\) as well. The derivative is positive on both sides of zero.
Step 3: Classify. The function increases on both sides. The derivative does not change sign at zero, so the graph does not turn there. The point \((0,0)\) is neither a local maximum nor a local minimum.
Check: The function values at nearby sample inputs agree with the sign test: \(f(-1)=-1\), \(f(0)=0\), and \(f(1)=1\). In particular, there are values below and above \(f(0)\) arbitrarily close to zero, which rules out both kinds of local extremum.
Worked Example: A Decreasing Cubic
Worked Example: A Decreasing Cubic
Classify the stationary number of \(g(x)=4-2(x+1)^3\).
Step 1: Differentiate. By the chain rule,
Step 2: Find where the derivative is zero.
The function value there is
The stationary point is \((-1,4)\).
Step 3: Test both sides. For \(x\neq -1\), \((x+1)^2>0\). Therefore
The derivative is negative both to the left and to the right of \(-1\). The function decreases on each side and does not turn at the stationary point. Thus \((-1,4)\) is neither a local maximum nor a local minimum.
Check: At one unit to either side, \(g(-2)=4-2(-1)^3=6\) and \(g(0)=4-2(1)^3=2\). These values are consistent with a decreasing function passing through \(g(-1)=4\): the value is higher on the left and lower on the right.
Worked Example: A Polynomial with More Than One Term
Worked Example: A Polynomial with More Than One Term
Determine whether \(h(x)=x^3+x^5\) has a local extremum at its stationary number \(x=0\).
Step 1: Differentiate and identify the stationary number.
At zero, \(h'(0)=3(0)^2+5(0)^4=0\), so zero is stationary. Also, \(h(0)=0^3+0^5=0\).
Step 2: Determine the sign on each side. For every \(x\neq 0\), \(x^2>0\), while \(3+5x^2>0\). Factoring the derivative gives
Thus \(h'\) is strictly positive immediately to the left and right of zero.
Step 3: Classify and check. The function increases on both sides, so it has no local extremum at zero. As a numerical check, \(h(-1)=-1-1=-2\), \(h(0)=0\), and \(h(1)=1+1=2\). These values agree with the sign analysis: the graph continues upward through the stationary point.
Here, as in the cubic example, the function’s derivative vanishes at the point but is positive on either side. The point has a horizontal tangent, yet the function does not change from increasing to decreasing or from decreasing to increasing.
A Common Mistake: Treating \(f'(c)=0\) as a Verdict
A common but incorrect calculation is:
The final conclusion does not follow from the displayed calculation. It confuses a necessary condition for an interior extremum of a differentiable function with a sufficient condition. Fermat’s theorem says that an interior local extremum of a differentiable function must have derivative zero; the converse is not guaranteed.
For the correction, test the derivative on both sides. Since \(3x^2>0\) for every \(x\neq 0\), the function is increasing on each side of zero. More directly, \(f(x)<f(0)\) for negative \(x\) and \(f(x)>f(0)\) for positive \(x\). The function takes both smaller and larger nearby values, so zero is neither a local minimum nor a local maximum.
A Reliable Classification Process
When a problem asks whether a stationary point is an extremum, use a sign test rather than guessing from the shape of a formula or from a single derivative value.
Differentiate and solve \(f'(x)=0\), checking that the candidate lies in the domain and is an interior point.
Use one interval immediately to the left of the candidate and another immediately to its right. Keep the intervals within the domain.
Find whether \(f'\) is positive or negative on each interval. A test value can help, but justify that the sign holds throughout the chosen interval.
If both sides are strictly positive, or both are strictly negative, the function continues in the same direction and the candidate is not a local extremum. Opposite signs indicate a change of direction.
This process is useful because stationary numbers are candidates for extrema, not guaranteed extrema. The sign test separates points where the function turns from points where its motion merely becomes momentarily horizontal.
Check Your Understanding
Use derivative signs on both sides of each stationary number to answer the questions.
- For \(p(x)=x^3\), what is the stationary number, and what are the signs of \(p'\) immediately to its left and right?
- For \(q(x)=-(x-2)^3\), find the stationary number and classify it using the derivative signs.
- Why does the equation \(f'(c)=0\) not, by itself, prove that \(c\) is a local maximum or minimum?
- State the Mean Value Theorem reasoning that shows a derivative strictly positive on both sides of a stationary number rules out a local extremum there.
- What is the error in concluding that a point must be a local minimum simply because the derivative is zero there?
