Why a Derivative Sign Controls Function Values
A derivative describes a function’s instantaneous rate of change. If that rate is positive throughout an interval, we expect the function’s values to rise as the input increases. If it is negative throughout, we expect the values to fall.
The Mean Value Theorem turns this intuition into a precise statement. It says that, between any two points, the average rate of change equals the derivative at some point between them. Thus, if the derivative has a fixed sign throughout the interval, every such average rate of change has that sign too.
This tutorial proves that connection and explains exactly what the hypotheses require. The conclusion applies not only to an entire domain, but also to any closed subinterval on which the hypotheses hold.
Monotonicity and the Mean Value Theorem
The non-strict and strict terms are not interchangeable. A nondecreasing function is allowed to have equal values at distinct inputs; a strictly increasing function is not. The same distinction applies to decreasing functions.
We use the Mean Value Theorem in the following form: if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), where \(a<b\), then there exists \(c\in(a,b)\) such that
The left side is the average rate of change from \(a\) to \(b\). The theorem guarantees that this average rate is the derivative at at least one interior point.
- If \(f'(x)\ge 0\) for every \(x\in(a,b)\), then \(f\) is nondecreasing on \([a,b]\).
- If \(f'(x)\le 0\) for every \(x\in(a,b)\), then \(f\) is nonincreasing on \([a,b]\).
- If \(f'(x)>0\) for every \(x\in(a,b)\), then \(f\) is strictly increasing on \([a,b]\).
- If \(f'(x)<0\) for every \(x\in(a,b)\), then \(f\) is strictly decreasing on \([a,b]\).
Proof of the Derivative-Sign Criterion
We prove the nondecreasing conclusion; the other cases follow by the same argument with the inequalities reversed or made strict. Take any two points \(x,y\in[a,b]\) with \(x<y\). The function is continuous on \([x,y]\) and differentiable on \((x,y)\), because these are contained in the interval where the hypotheses hold.
By the Mean Value Theorem, some \(c\in(x,y)\) satisfies
Since \(y-x>0\) and \(f'(c)\ge 0\), multiplying both sides by \(y-x\) gives
Therefore \(f(x)\le f(y)\). The points \(x\) and \(y\) were arbitrary, so this inequality holds for every pair with \(x<y\); that is exactly the definition of being nondecreasing on \([a,b]\).
If \(f'(x)>0\) throughout \((a,b)\), then the point \(c\) supplied by the theorem has \(f'(c)>0\). Since \(y-x>0\), the same equation gives \(f(y)-f(x)>0\), which proves strict increase. If \(f'(x)\le0\), the product \((y-x)f'(c)\) is nonpositive, so \(f(y)\le f(x)\). If \(f'(x)<0\), it is strictly negative, so \(f(y)<f(x)\). This proves all four statements.
Applying the Result on a Closed Interval
To use the theorem on \([a,b]\), check its hypotheses on that interval: \(f\) must be continuous at every point of \([a,b]\), differentiable at every point of \((a,b)\), and its derivative must have the required sign throughout the open interval. The derivative condition need not be checked at the endpoints, where the ordinary two-sided derivative may not be defined.
The theorem then compares any two function values in the interval, including values at the endpoints. In particular, if \(f'\ge0\) on \((a,b)\), then \(f(a)\le f(x)\le f(b)\) for every \(x\in[a,b]\). This follows by applying nondecrease first to the pair \(a,x\), then to the pair \(x,b\). The corresponding inequalities reverse when \(f'\le0\).
Worked Example: A Positive Derivative on a Closed Interval
Let \(f(x)=x^2\) on \([1,3]\). Determine how the function behaves and compare its endpoint values.
First, \(f\) is a polynomial, so it is continuous on \([1,3]\) and differentiable on \((1,3)\). Its derivative is
For every \(x\in(1,3)\), \(x>0\), so \(f'(x)=2x>0\). The strict version of the theorem shows that \(f\) is strictly increasing on \([1,3]\).
For the endpoint values,
The endpoint difference is \(9-1=8>0\), consistent with strict increase. As a direct check of the Mean Value Theorem for these endpoints, the average rate of change is
The derivative equals \(4\) at \(c=2\), since \(f'(2)=2(2)=4\), and \(2\in(1,3)\), as the theorem requires.
Worked Example: A Nonnegative Derivative That Vanishes
Let \(f(x)=x^3\) on \([-1,2]\). Use the derivative-sign criterion, and explain what it does and does not establish.
The function is a polynomial, so it is continuous on the closed interval and differentiable in its interior. Its derivative is
Because \(x^2\ge0\) for every real \(x\), we have \(f'(x)\ge0\) throughout \((-1,2)\). The theorem therefore proves that \(f\) is nondecreasing on \([-1,2]\).
At the endpoints,
Thus \(f(2)-f(-1)=8-(-1)=9>0\), in agreement with the theorem. The derivative is zero at \(x=0\), so the strict-derivative condition \(f'(x)>0\) does not apply on the whole interval. That condition is sufficient for strict increase, but it is not necessary.
In this example strict increase can also be verified directly. For any \(x<y\),
Here \(y-x>0\). Also,
This sum of nonnegative terms can be zero only if \(x=0\) and \(y=0\), which is impossible when \(x<y\). Therefore \(y^2+xy+x^2>0\), so \(y^3-x^3>0\). This proves strict increase even though the derivative is not positive at every interior point.
Worked Example: A Negative Derivative on a Closed Interval
Let \(f(x)=1/x\) on \([1,4]\). Determine its monotonicity and compare its endpoint values.
Since \(0\notin[1,4]\), the function is continuous on \([1,4]\) and differentiable on \((1,4)\). Differentiating gives
For each \(x\in(1,4)\), \(x^2>0\), so \(-1/x^2<0\). The theorem shows that \(f\) is strictly decreasing on \([1,4]\).
The endpoint values are
Since \(1>1/4\), the endpoint comparison agrees with strict decrease. The theorem also compares every pair of points in the interval, not just these endpoints.
What the Criteria Do Not Say
A derivative sign gives a reliable conclusion only on an interval where the sign condition holds throughout. If the derivative changes sign, the theorem does not establish monotonicity across the entire interval. This does not mean the theorem has failed; its hypothesis is simply unavailable there. A smaller interval on which the sign is fixed may still be suitable.
Also distinguish “the theorem guarantees” from “the function must.” A strictly positive derivative guarantees strict increase. A nonnegative derivative guarantees nondecrease. But a function can be strictly increasing even when its derivative is zero at some points, as the cubic example showed. A zero derivative at a point alone does not establish either increase or decrease.
There is a useful converse when the derivative exists. If \(f\) is nondecreasing on an interval and differentiable at an interior point \(x\), then \(f'(x)\ge0\). For sufficiently small \(h>0\), nondecrease gives \(f(x+h)-f(x)\ge0\), so
Taking the limit as \(h\) approaches zero from the right yields \(f'(x)\ge0\), because the two-sided derivative exists and therefore agrees with that one-sided limit. Similarly, a nonincreasing differentiable function has \(f'(x)\le0\). This converse concerns non-strict monotonicity; strict increase does not require a strictly positive derivative at every point.
A Common Mistake: Extending a Conclusion Past Its Hypotheses
Worked Example: A Derivative That Changes Sign
Consider \(f(x)=x^2\) on \([-2,1]\). A mistaken argument might say: “\(f'(x)=2x<0\) for negative \(x\), so \(f\) is decreasing on the whole interval.” This uses the derivative sign only on part of the interval.
In fact, \(f'(x)=2x\) is negative on \((-2,0)\), is zero at \(0\), and is positive on \((0,1)\). Thus there is no single nonpositive derivative sign throughout \((-2,1)\), and the decreasing criterion cannot be applied to the whole interval.
The claimed conclusion is also contradicted by values within the interval:
A decreasing function would have to satisfy \(f(-1)\ge f(0)\), which this pair does; to test the claimed decrease further, compare \(0<1\): \(f(0)=0<1=f(1)\), contradicting the requirement \(f(0)\ge f(1)\). The correction is to apply the theorem only where its sign condition holds: \(f\) is decreasing on \([-2,0]\) and increasing on \([0,1]\). The derivative sign does not support either conclusion across the entire interval.
Using the Theorem Carefully
State the interval on which you want to compare function values. Check whether the derivative sign is fixed on that entire interval.
For a closed interval \([a,b]\), verify continuity on \([a,b]\) and differentiability on \((a,b)\).
Use \(f'\ge0\) for nondecrease, \(f'\le0\) for nonincrease, \(f'>0\) for strict increase, or \(f'<0\) for strict decrease.
For any \(x<y\) in the interval, the Mean Value Theorem gives a point \(c\in(x,y)\) whose derivative determines the sign of \(f(y)-f(x)\).
The essential logic is that each pair \(x<y\) produces its own intermediate point \(c\). A derivative sign that holds everywhere in the relevant interval controls every such pair, which is why it determines monotonicity rather than merely describing behavior at one point.
Check Your Understanding
For each question, identify the relevant hypotheses and explain what conclusion the derivative-sign criterion permits.
- If \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f'(x)\ge0\) throughout \((a,b)\), what monotonicity conclusion follows?
- Why does the Mean Value Theorem require differentiability only between the two points being compared, rather than at the endpoints of the full interval?
- If \(f'(x)>0\) at all but one interior point and \(f'(x)=0\) at that point, does the strict-derivative condition in the theorem apply as stated? What conclusion does the nonnegative condition give?
- For \(f(x)=1/x\) on \([1,4]\), verify the derivative sign and state the order of \(f(1)\) and \(f(4)\).
- If a derivative is negative on one part of an interval and positive on another, why can neither fixed-sign criterion be applied to the whole interval?
