Difference Quotients Preserve the Sign
For a nondecreasing function, moving to the right cannot lower the function value, and moving to the left cannot raise it. The derivative records these comparisons through difference quotients. The sign of the denominator matters: for a negative increment, both the change in the function and the increment are nonpositive, so their quotient is still nonnegative.
Throughout, let \(I\subseteq\mathbb{R}\) be an interval with more than one point, and let \(f:I\to\mathbb{R}\) be real-valued. “Nondecreasing” means that \(f(s)\le f(t)\) whenever \(s,t\in I\) and \(s<t\); “strictly increasing” means \(f(s)<f(t)\) whenever \(s<t\). Derivatives below are ordinary two-sided derivatives at interior points of \(I\). A finite derivative means that the difference quotient has a limit in \(\mathbb{R}\), not an infinite limit. All statements are pointwise.
Proof. For every sufficiently small \(h>0\), both \(x\) and \(x+h\) belong to \(I\), and nondecrease gives \(f(x+h)-f(x)\ge0\). Since \(h>0\), the quotient \(\bigl(f(x+h)-f(x)\bigr)/h\) is nonnegative. For every sufficiently small \(h<0\), nondecrease gives \(f(x+h)\le f(x)\), so \(f(x+h)-f(x)\le0\). Dividing this nonpositive numerator by \(h<0\) again gives a nonnegative quotient. Thus the difference quotient is nonnegative for every sufficiently small nonzero \(h\). Its limit exists and equals \(f'(x)\); a limit of nonnegative real numbers is nonnegative. Therefore \(f'(x)\ge0\). \(\square\)
The argument does not require continuity of \(f\) on the whole interval or boundedness of \(I\). It does require that \(x\) be an interior point, so that increments of both signs are available, and that the ordinary finite derivative exist. At an endpoint, a one-sided derivative requires a separate statement and proof using only increments that stay in \(I\).
What a Positive Derivative Says Locally
The sign theorem gives a necessary condition on the derivative. A strictly positive derivative gives more: sufficiently close to the point, values to the right are strictly larger, and values to the left are strictly smaller. These are different inequalities on the two sides, not a claim that every nearby value exceeds \(f(x)\).
Proof. By the definition of derivative, for \(\varepsilon=d/2>0\) there is a \(\delta>0\) such that whenever \(0<|h|<\delta\) and \(x+h\in I\),
This implies that the quotient is greater than \(d/2\), hence positive. If \(0<h<\delta\), multiplying the positive quotient by \(h>0\) gives \(f(x+h)-f(x)>0\). If \(-\delta<h<0\), multiplying it by \(h<0\) reverses the inequality and gives \(f(x+h)-f(x)<0\). Both asserted comparisons follow. \(\square\)
Worked Example: A Positive Derivative on Both Sides
On \(I=\mathbb{R}\), consider \(f(t)=2t+3\) at \(x=1\). This function is strictly increasing because for \(s<t\), \(f(t)-f(s)=2(t-s)>0\). Its derivative can be computed directly: for every \(h\ne0\),
Thus \(f'(1)=2>0\). The proposition applies, and in fact the local comparisons hold for every allowed \(h\ne0\): \(f(1+h)-f(1)=2h\), which is positive when \(h>0\) and negative when \(h<0\). For instance, at \(h=1/4\), \(f(5/4)=11/2>5=f(1)\); at \(h=-1/4\), \(f(3/4)=9/2<5\). Check: the corresponding difference quotients are \((11/2-5)/(1/4)=2\) and \((9/2-5)/(-1/4)=2\), agreeing with the derivative.
Strict Increase Does Not Require a Positive Derivative
Strict increase is a comparison between values at distinct points. A positive derivative is a stronger, local first-order condition. Strict increase alone does not force the derivative, when it exists, to be positive: it can be zero. The nonnegativity theorem remains correct in this case, but its conclusion does not characterize strict increase.
Worked Example: Strict Increase with Zero Derivative
Let \(f(t)=t^3\) on \(I=\mathbb{R}\), and examine \(x=0\). For \(s<t\), factor the difference:
The first factor is positive. The second is nonnegative because \(t^2+ts+s^2=(t+s/2)^2+3s^2/4\). It can be zero only if \(s=t=0\), which is impossible when \(s<t\). Thus \(f(t)-f(s)>0\), and \(f\) is strictly increasing.
At zero, \(f(0)=0\), and for every \(h\ne0\),
As \(h\to0\), \(h^2\to0\), so \(f'(0)=0\). Yet \(f(h)>f(0)\) for every \(h>0\), and \(f(h)<f(0)\) for every \(h<0\). Check: at \(h=1/2\), the value is \(f(1/2)=1/8>0\), and the quotient is \((1/8)/(1/2)=1/4=(1/2)^2\). At \(h=-1/2\), the value is \(-1/8<0\), and the quotient is \((-1/8)/(-1/2)=1/4\). The derivative is zero even though the function is strictly increasing on both sides.
More generally, if \(f'(x)=0\), the definition says that for each \(\varepsilon>0\), sufficiently small nonzero \(h\) satisfy \(\left|f(x+h)-f(x)\right|<\varepsilon|h|\). This controls the size of the change relative to \(|h|\); it does not say that the change is zero. The cubic example makes that distinction explicit.
Flat Intervals and Their Endpoints
A function is flat on an interval when it is constant there. At every interior point of such an interval, its derivative exists and is zero: all sufficiently small increments remain in the interval, so the difference quotient is exactly zero. At an endpoint, increments may leave the flat part, and the derivative need not exist. The endpoint must be checked using both sides if an ordinary derivative is claimed.
Worked Example: A Monotone Function with a Flat Interval
Define \(f:[-2,2]\to\mathbb{R}\) by
On \([-2,-1)\), the function increases with slope \(1\), and its values are negative. On \([-1,1]\), its value is \(0\). On \((1,2]\), it increases with slope \(1\), and its values are positive. These ranges and the increase within each nonconstant piece show that \(f(s)\le f(t)\) whenever \(s<t\); thus \(f\) is nondecreasing. It is not strictly increasing, since \(f(0)=f(1/2)=0\).
For any \(x\in(-1,1)\), choose \(\delta=\min\{x+1,1-x\}>0\). Whenever \(|h|<\delta\), both \(x\) and \(x+h\) lie in \((-1,1)\), so \(f(x+h)=f(x)=0\). Therefore the difference quotient is zero for every such nonzero \(h\), and \(f'(x)=0\).
At \(x=1\), the left difference quotient is \(0\) for \(-1<h<0\), because \(1+h\in(0,1)\). For \(0<h\le1\), \(f(1+h)=h\), so the right difference quotient is \(h/h=1\). The left and right limits are \(0\) and \(1\), respectively, so \(f'(1)\) does not exist. At \(x=-1\), for \(-1<h<0\), \(f(-1+h)=h\), giving quotient \(h/h=1\); for \(0<h<1\), \(f(-1+h)=0\), giving quotient \(0/h=0\). Again the two one-sided limits differ, so \(f'(-1)\) does not exist. Check: at \(x=0\), every \(|h|<1\) stays in the flat interval and gives quotient \(0\); at \(x=1\), \(h=-1/2\) gives quotient \(0\), while \(h=1/2\) gives quotient \(1\). These computations confirm the interior derivative and the endpoint failures.
A Sign Pitfall and the Scope of the Conclusion
A common error is to reason that the numerator \(f(x+h)-f(x)\) is nonnegative for both positive and negative \(h\). That is not what nondecrease says. The correct comparison changes direction when the increment changes sign.
For a concrete check, take \(f(t)=t\), \(x=0\), and \(h=-1/10\). Then \(f(x+h)-f(x)=-1/10\), not a nonnegative number. The quotient is nevertheless positive:
The correction is to use \(f(x+h)\le f(x)\) when \(h<0\), and then divide the nonpositive difference by the negative increment. This yields a nonnegative quotient, just as the positive-increment case does.
The results here give pointwise implications, not converses. For a nondecreasing function, a finite derivative cannot be negative. If that derivative is positive, the function has the stated strict comparisons sufficiently close to the point. If the derivative is zero, the function may still be strictly increasing, as \(t^3\) shows, or may be locally constant, as on the interior of the plateau. The derivative sign alone does not decide between those behaviors.
Check Your Understanding
For each question, track the sign of the increment as well as the sign of the function change.
- Let \(f\) be nondecreasing on an interval and differentiable at an interior point \(x\). Why is the difference quotient nonnegative for \(h<0\)?
- If \(f'(x)=d>0\), what strict comparison holds for \(f(x+h)\) when \(h>0\) is sufficiently small? What comparison holds when \(h<0\)?
- Verify directly that \(f(t)=t^3\) is strictly increasing, yet \(f'(0)=0\).
- For the flat-interval example, compute the left and right difference-quotient limits at \(x=-1\).
- Does \(f'(x)=0\) imply that \(f\) is constant near \(x\)? Give an example or explain why not.