From a Derivative Sign to an Order of Function Values
A derivative describes local change: its sign indicates whether a function is increasing or decreasing at a point. The Mean Value Theorem connects that local information to a comparison between function values at two different points. If the derivative never has the wrong sign between those points, the function cannot move in the opposite direction overall.
The conclusion is a statement about every pair of points in an interval, not just about a particular graph or a few sample values. The conditions matter: continuity on the segment between the points and differentiability in its interior are what allow the Mean Value Theorem to make the connection.
Nondecreasing and Nonincreasing Functions
These definitions permit equality. A nondecreasing function may stay constant over part of its domain; it does not have to rise at every step. Likewise, a nonincreasing function may remain constant. The derivative conditions below therefore use “nonnegative” and “nonpositive,” rather than requiring a strictly positive or strictly negative derivative.
The Monotonicity Theorem
For an open interval, “every interior point of \(I\)” means every point of \(I\). If \(I\) has endpoints, the derivative sign is needed only at points inside the interval: the proof applies the theorem to the segment between any two distinct points of \(I\), and its Mean Value Theorem point lies strictly between them.
Why the Theorem Works
Take any two points \(x_1,x_2\in I\) with \(x_1<x_2\). The function is continuous on \([x_1,x_2]\) and differentiable on \((x_1,x_2)\), so the Mean Value Theorem applies. It gives a point \(c\in(x_1,x_2)\) for which the change in function value equals the interval length multiplied by the derivative at \(c\):
Since \(x_2-x_1>0\), the sign of the difference \(f(x_2)-f(x_1)\) is determined by the sign of \(f'(c)\). If the derivative is nonnegative throughout the interval, then \(f'(c)\geq 0\), so \(f(x_2)-f(x_1)\geq 0\). Thus \(f(x_1)\leq f(x_2)\). The choice of points was arbitrary, so this holds for every ordered pair in \(I\). The nonincreasing conclusion follows by the same reasoning with the inequalities reversed.
Take \(x_1<x_2\) in the interval. The goal is to compare \(f(x_1)\) and \(f(x_2)\).
Continuity on \([x_1,x_2]\) and differentiability on \((x_1,x_2)\) give a \(c\in(x_1,x_2)\) with \(f(x_2)-f(x_1)=f'(c)(x_2-x_1)\).
The interval length \(x_2-x_1\) is positive. Therefore \(f'(c)\geq 0\) forces \(f(x_2)-f(x_1)\geq 0\), while \(f'(c)\leq 0\) forces \(f(x_2)-f(x_1)\leq 0\).
Because the two points were arbitrary, the required order relation holds for every pair in \(I\).
Worked Examples
Worked Example: A Nonnegative Derivative on a Closed Interval
Show that \(f(x)=x^2\) is nondecreasing on \([0,3]\), then check the comparison between \(x=1\) and \(x=3\) using the Mean Value Theorem.
The function is a polynomial, so it is continuous on \([0,3]\) and differentiable on \((0,3)\). Its derivative is
For every \(x\in(0,3)\), \(x>0\), and therefore \(2x>0\), which in particular means \(f'(x)\geq 0\). The theorem gives that \(f\) is nondecreasing on \([0,3]\).
For the specified pair, \(f(1)=1^2=1\) and \(f(3)=3^2=9\), so \(f(3)-f(1)=9-1=8\). The Mean Value Theorem gives a \(c\in(1,3)\) satisfying
Since \(f'(c)=2c\), this requires \(2c=4\), hence \(c=2\), which lies in \((1,3)\). Checking the theorem’s equation, \(f'(2)(3-1)=4\cdot2=8=f(3)-f(1)\). The computed values agree with the nondecreasing conclusion.
Worked Example: A Nonpositive Derivative
Show that \(g(x)=-x^3\) is nonincreasing on \(\mathbb{R}\), and check the comparison between \(x=0\) and \(x=2\).
The polynomial \(g\) is continuous and differentiable everywhere. Its derivative is
Because \(x^2\geq 0\) for every real \(x\), we have \(-3x^2\leq 0\) everywhere. The theorem therefore proves that \(g\) is nonincreasing on \(\mathbb{R}\).
At the chosen points, \(g(0)=-(0)^3=0\) and \(g(2)=-(2)^3=-8\). Thus \(g(2)-g(0)=-8\). The Mean Value Theorem guarantees a \(c\in(0,2)\) for which
Substituting the derivative gives \(-3c^2=-4\), so \(c^2=4/3\). The solution in \((0,2)\) is \(c=2/\sqrt{3}\), which is positive and less than \(2\). Finally, \(g'(c)(2-0)=(-4)(2)=-8=g(2)-g(0)\). The endpoint values are in the order required for a nonincreasing function: \(0\geq -8\).
Worked Example: A Derivative That Vanishes Inside the Interval
Show that \(h(x)=x^3/3\) is nondecreasing on \([-1,2]\), even though its derivative is zero at an interior point. Check the pair \(x=-1\) and \(x=2\).
The function is a polynomial, and hence is continuous on \([-1,2]\) and differentiable on \((-1,2)\). Its derivative is
Since \(x^2\geq 0\) for every real \(x\), the theorem proves that \(h\) is nondecreasing on \([-1,2]\). In particular, \(h'(0)=0\), but this does not invalidate the condition: the derivative is still nonnegative at \(0\) and everywhere else in the interval.
For the endpoint check,
The change is \(h(2)-h(-1)=8/3-(-1/3)=9/3=3\), and the interval length is \(2-(-1)=3\). The Mean Value Theorem therefore requires \(h'(c)=3/3=1\) for some \(c\in(-1,2)\). Since \(h'(c)=c^2\), we need \(c^2=1\); both \(c=-1\) and \(c=1\) solve the equation, but only \(c=1\) lies in the open interval \((-1,2)\). Checking, \(h'(1)(2-(-1))=1\cdot3=3=h(2)-h(-1)\). The endpoint values are ordered correctly despite the zero derivative at an interior point.
Why the Conditions Must Hold Throughout the Interval
The Mean Value Theorem’s point \(c\) can be anywhere strictly between the two selected points. A derivative sign at only one endpoint, or at a few sampled points, says nothing sufficient about the derivative at that \(c\). To guarantee the comparison for every pair of points, the sign condition must hold throughout the interval’s interior.
Continuity and differentiability are also essential hypotheses of the Mean Value Theorem. If either fails on a segment, this argument cannot be used on that segment. A derivative calculation alone does not establish that the theorem applies; check the regularity conditions on each closed segment being considered.
Common Mistake: Confusing a Nonnegative Square with a Nonnegative Derivative
A frequent error is to claim that \(f(x)=x^2\) is nondecreasing on \([-1,1]\) because “squares are nonnegative.” That argument checks the values of \(f\), not the sign of its derivative. In fact,
which is negative when \(-1<x<0\) and positive when \(0<x<1\). The derivative is not nonnegative throughout \((-1,1)\), so the monotonicity theorem does not prove that \(f\) is nondecreasing on the whole interval. Indeed, \(f(-1)=1\) and \(f(0)=0\); since \(-1<0\) but \(1\not\leq 0\), the definition directly shows that \(f\) is not nondecreasing there.
The correction is to check the derivative’s sign on the full interval before applying the theorem. On \([0,1]\), \(2x\geq 0\), so \(x^2\) is nondecreasing. On \([-1,0]\), \(2x\leq 0\), so \(x^2\) is nonincreasing. Neither conclusion extends across the whole interval as a single monotonicity claim.
Using the Result Reliably
When proving monotonicity from a derivative, organize the reasoning around the theorem’s hypotheses and conclusion:
- State the interval on which the claim is being made.
- Verify continuity on the closed segment between any two points and differentiability in its interior.
- Compute the derivative and establish its sign at every interior point of the interval.
- Invoke the Mean Value Theorem for arbitrary \(x_1<x_2\) in the interval.
- Use \(x_2-x_1>0\) to translate the derivative sign into the corresponding comparison of function values.
The method is useful because it converts a pointwise derivative condition into a global order statement on an interval. The essential step is not merely calculating a derivative; it is ensuring that the sign condition and the Mean Value Theorem’s hypotheses apply wherever the two chosen points lie.
Check Your Understanding
Answer these questions using the definitions and theorem above.
- State the continuity and differentiability conditions needed to apply the Mean Value Theorem between two points \(x_1<x_2\).
- If \(f'(x)\geq 0\) at every interior point of an interval, what inequality must hold between \(f(x_1)\) and \(f(x_2)\) when \(x_1<x_2\)?
- Why does a derivative equal to zero at one interior point not prevent the theorem from proving nondecreasing behavior?
- For \(p(x)=-x^2\) on \([0,2]\), compute \(p'(x)\) and determine whether the theorem proves that \(p\) is nondecreasing or nonincreasing.
- What is wrong with concluding that \(x^2\) is nondecreasing on \([-1,1]\) merely because \(x^2\geq 0\)?
