Why a Derivative Sign Can Rule Out an Extremum
A local maximum or minimum is a point where a function’s values turn around, at least in a small neighborhood. A derivative that keeps the same strict sign throughout an interval tells us that no such turn can occur there: the function moves in just one direction across the whole interval.
For an AP Calculus justification, the important step is to connect the sign of the derivative to a statement about function values. The Mean Value Theorem supplies that connection. Once strict monotonicity has been established, it rules out every interior local extremum in the interval—not merely the points where a derivative test happens to be inconclusive.
Definitions and Conditions
A function \(f\) is strictly increasing on an interval \(I\) if, whenever \(x_1,x_2\in I\) and \(x_1<x_2\), we have \(f(x_1)<f(x_2)\). It is strictly decreasing on \(I\) if those same inputs satisfy \(f(x_1)>f(x_2)\).
A point \(c\) is an interior local maximum of \(f\) if \(c\) is an interior point of the domain and there is a neighborhood of \(c\) in which \(f(c)\geq f(x)\) for all domain values \(x\). An interior local minimum is defined by \(f(c)\leq f(x)\) throughout some neighborhood. The inequalities may be non-strict, so the definition includes local extrema that are not strict.
Here, saying \(f\) is differentiable on an interval means that it is differentiable at every point in the interval. To apply the Mean Value Theorem between two points of the interval, \(f\) must be continuous on the closed segment between them and differentiable on the segment’s interior. Differentiability on the interval ensures these conditions for any such segment.
Why the Theorem Works
The Mean Value Theorem states that if \(f\) is continuous on \([x_1,x_2]\) and differentiable on \((x_1,x_2)\), then there is a number \(c\in(x_1,x_2)\) such that
The denominator is positive when \(x_1<x_2\). Thus, the sign of the derivative at the intermediate point \(c\) determines the sign of the change \(f(x_2)-f(x_1)\). If the derivative is positive everywhere, that change must be positive; if the derivative is negative everywhere, the change must be negative.
Take arbitrary \(x_1,x_2\in I\) with \(x_1<x_2\). The whole segment \([x_1,x_2]\) lies in \(I\), since \(I\) is an interval.
There is \(c\in(x_1,x_2)\) for which \(f(x_2)-f(x_1)=f'(c)(x_2-x_1)\).
If \(f'(x)>0\) throughout \(I\), then \(f'(c)>0\), and \(x_2-x_1>0\), so \(f(x_2)-f(x_1)>0\). Hence \(f(x_1)<f(x_2)\). If \(f'(x)<0\), the same equation gives \(f(x_2)-f(x_1)<0\), hence \(f(x_1)>f(x_2)\).
At any interior point \(c\), there are points of \(I\) just to its left and just to its right. In the increasing case, nearby left values are smaller than \(f(c)\) and nearby right values are larger. Thus \(c\) can be neither a local maximum nor a local minimum. In the decreasing case, nearby left values are larger and nearby right values are smaller, with the same conclusion.
This proves both claims. Strict monotonicity compares every pair of ordered inputs; it is stronger than observing that the function rises or falls at a few selected points.
Worked Examples
Worked Example: A Positive Derivative Throughout an Interval
Show that \(f(x)=x^3+2x\) has no interior local extrema on \((-2,3)\).
Step 1: Differentiate.
For every real \(x\), \(x^2\geq 0\), so \(3x^2\geq 0\) and \(3x^2+2\geq 2>0\). In particular, \(f'(x)>0\) throughout \((-2,3)\).
Step 2: Apply the theorem. The function is differentiable on the interval, and its derivative is strictly positive there. Therefore \(f\) is strictly increasing on \((-2,3)\), so it has no interior local maximum or minimum in that interval.
Check: For example, \(f(-1)=(-1)^3+2(-1)=-1-2=-3\), while \(f(1)=1^3+2(1)=1+2=3\). Since \(-1<1\), the values \(-3<3\) agree with the increasing conclusion. This check illustrates the result; the Mean Value Theorem argument establishes it for every pair of inputs.
Worked Example: A Negative Derivative Throughout an Interval
Show that \(g(x)=\frac{1}{x}\) has no interior local extrema on \((0,\infty)\).
Step 1: Check the domain and differentiate. The function is defined and differentiable for every \(x>0\). Its derivative is
Because \(x>0\), we have \(x^2>0\), so \(-1/x^2<0\) throughout \((0,\infty)\).
Step 2: Apply the theorem. The derivative is strictly negative throughout the interval, so \(g\) is strictly decreasing there. It follows that \(g\) has no interior local maximum or minimum on \((0,\infty)\).
Check: Take \(x_1=1\) and \(x_2=2\). Then \(x_1<x_2\), \(g(1)=1\), and \(g(2)=1/2\), so \(g(1)>g(2)\), as required for a decreasing function. The general conclusion comes from the derivative sign and the theorem, not from this one comparison.
Worked Example: Factoring to Establish the Sign
Show that \(h(x)=x^3-6x^2+9x\) has no interior local extrema on \((1,3)\).
Step 1: Differentiate and factor.
For every \(x\in(1,3)\), \(x-1>0\) and \(x-3<0\). Their product is negative, and multiplication by \(3>0\) preserves that sign. Therefore \(h'(x)<0\) at every point of \((1,3)\).
Step 2: Conclude. The function is differentiable on \((1,3)\), so the derivative sign theorem shows that \(h\) is strictly decreasing there. It has no interior local maximum or minimum on that interval.
Check: Rewrite \(h(x)=x(x-3)^2\). Then \(h(3/2)=(3/2)(-3/2)^2=(3/2)(9/4)=27/8\), and \(h(5/2)=(5/2)(-1/2)^2=(5/2)(1/4)=5/8\). Since \(3/2<5/2\) and \(27/8>5/8\), these exact values agree with the decreasing conclusion.
Use the Whole Interval, Not One Point
A derivative’s sign at a single point does not establish monotonicity on an interval. The theorem requires the stated strict sign at every point of the interval under consideration. This is why a sign analysis must cover the entire interval where the conclusion is being claimed.
For example, let \(q(x)=x^2\) on \((-1,1)\). A tempting but incorrect argument is: “\(q'(1/2)=1>0\), so \(q\) is increasing on \((-1,1)\), and it has no local extrema there.” The derivative calculation is
But the calculation only gives the derivative at \(1/2\), not throughout \((-1,1)\). In fact, \(q'(x)=2x<0\) when \(-1<x<0\), and \(q'(x)=2x>0\) when \(0<x<1\). Also, \(q(0)=0\), while \(q(x)=x^2>0\) for every nonzero \(x\) near \(0\). Thus \(0\) is an interior local minimum, contradicting the incorrect conclusion.
What the Conclusion Does—and Does Not—Say
If \(f\) is strictly increasing on an interval, then no interior point of that interval can be a local extremum. The same is true if \(f\) is strictly decreasing. The word interior matters: an interval endpoint has no interval points on one side, so this argument does not rule out an endpoint extremum.
The conclusion is also limited to the interval where the derivative sign has been established. A function might have an extremum elsewhere in its domain. In the example \(q(x)=x^2\), for instance, the derivative is positive on \((0,1)\), so the function is strictly increasing there; that fact does not describe what happens on \((-1,0)\), where the derivative is negative.
When writing a justification, state the interval, the derivative sign on that whole interval, the resulting monotonicity, and the consequence for interior extrema. For example: “Since \(f'(x)<0\) for every \(x\in(a,b)\), the Mean Value Theorem implies that \(f\) is strictly decreasing on \((a,b)\). Therefore \(f\) has no interior local maximum or minimum on that interval.” This makes the evidence and conclusion explicit.
Check Your Understanding
For each question, give a reasoned answer rather than relying on a graph or a few sample values.
- If \(f\) is differentiable on \((a,b)\) and \(f'(x)>0\) for every \(x\in(a,b)\), what monotonicity conclusion follows?
- Which theorem connects the derivative sign to the difference \(f(x_2)-f(x_1)\) for \(x_1<x_2\)?
- Why does a strictly increasing function have neither a local maximum nor a local minimum at an interior point?
- Suppose \(f'(2)>0\). Is that enough to conclude that \(f\) is increasing on \((1,3)\)? Explain.
- Does a strict derivative sign on an open interval rule out a local extremum at an endpoint of its closure? Explain why or why not.
