From a Derivative Sign to a Comparison
A derivative describes a function’s instantaneous rate of change. If the derivative is positive throughout an interval, the function rises locally there. The Mean Value Theorem lets us turn that local information into a precise comparison between any two points in the interval. This is why a derivative sign can establish monotonicity over an entire interval, rather than merely suggest it from a graph.
The key is to apply the Mean Value Theorem to the segment between two arbitrary points. Its conclusion relates the difference in function values to a derivative at some point between them.
The Derivative-Sign Theorem
The Mean Value Theorem states that if \(f\) is continuous on a closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then there is at least one \(c\in(a,b)\) such that
The number \(c\) may depend on the chosen endpoints \(a\) and \(b\). For monotonicity, we do not need to find it exactly. We need only know that it lies between the endpoints, where the derivative sign is controlled.
For an open interval, the continuity and differentiability conditions mean they hold at the points of that interval. For a closed interval, continuity is required at its endpoints, while differentiability is required at interior points; the derivative need not exist at an endpoint. The theorem works on unbounded intervals as well, because any two points chosen from such an interval still determine a finite closed segment.
Why the Theorem Works
Here is the proof for a positive derivative. Choose 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 \(c\in(x_1,x_2)\) satisfying
Since \(c\in I\), the assumption gives \(f'(c)>0\). Also \(x_2-x_1>0\). The product of these two positive quantities is positive, so
The points were arbitrary, so this comparison holds for every ordered pair in \(I\). That is exactly the definition of strictly increasing.
For a negative derivative, the same equation has \(f'(c)<0\) and \(x_2-x_1>0\). Thus \(f(x_2)-f(x_1)<0\), which means \(f(x_2)<f(x_1)\): the function is strictly decreasing. If the derivative is only nonnegative, the product is at least zero, yielding \(f(x_1)\leq f(x_2)\); this proves the nondecreasing conclusion. The nonincreasing case follows in the same way from a nonpositive derivative.
Take arbitrary \(x_1<x_2\) from the interval whose monotonicity is being studied.
Verify continuity on \([x_1,x_2]\) and differentiability on \((x_1,x_2)\).
Use \(f(x_2)-f(x_1)=f'(c)(x_2-x_1)\) for some \(c\) between the chosen points.
Combine the sign of \(f'(c)\) with \(x_2-x_1>0\) to compare the function values.
Worked Examples
Worked Example: Proving Increase on an Open Interval
Show that \(f(x)=x^3+3x\) is strictly increasing on \((-2,2)\). Then check the comparison between \(-1\) and \(1\).
First differentiate:
Since \(x^2\geq 0\), we have \(x^2+1\geq 1\), so \(f'(x)\geq 3>0\) for every \(x\in(-2,2)\). The function is a polynomial, hence continuous and differentiable everywhere. By the theorem, \(f\) is strictly increasing on \((-2,2)\).
For the requested check, \(-1<1\), and
Thus \(f(1)-f(-1)=4-(-4)=8>0\). The Mean Value Theorem also predicts a point \(c\in(-1,1)\) with
Solving \(3c^2+3=4\) gives \(3c^2=1\), \(c^2=\frac13\), and \(c=\pm\frac{1}{\sqrt3}\), both of which lie in \((-1,1)\). Substitution checks either value: \(f'(c)=3(\frac13)+3=4\), the required secant slope.
Worked Example: Proving Decrease on a Closed Interval
Show that \(g(x)=\frac{1}{x+1}\) is strictly decreasing on \([0,3]\). Check the Mean Value Theorem for the pair \(0,2\).
The denominator \(x+1\) is nonzero on \([0,3]\). The function is continuous there and differentiable in the interior. Its derivative is
Because \((x+1)^2>0\) on the interval, \(g'(x)<0\). The theorem therefore proves that \(g\) is strictly decreasing on \([0,3]\).
For \(a=0\) and \(b=2\), compute
The secant slope is \(\frac{-2/3}{2-0}=-\frac13\). The Mean Value Theorem predicts \(c\in(0,2)\) with \(g'(c)=-\frac13\). Solving,
Since \(c+1>0\), \(c+1=\sqrt3\), so \(c=\sqrt3-1\). This lies in \((0,2)\), since \(1<\sqrt3<3\). Substitution gives \(g'(c)=-1/(\sqrt3)^2=-\frac13\), confirming the required slope. The endpoint values also agree with decrease: \(g(2)=\frac13<1=g(0)\).
Worked Example: An Endpoint Where the Derivative Is Undefined
Show that \(h(x)=\sqrt{x+1}\) is strictly increasing on \([-1,3]\), even though its derivative is not defined at the left endpoint. Check the pair \(-1,3\).
The square root function is continuous on \([-1,3]\). For interior points \(x\in(-1,3)\), differentiation gives
The derivative is positive because \(x+1>0\) at every interior point. The derivative is undefined at \(x=-1\), but this does not prevent applying the Mean Value Theorem: it requires differentiability only on the open interval between the chosen endpoints. Therefore \(h\) is strictly increasing on \([-1,3]\).
For the endpoint pair, \(h(-1)=\sqrt0=0\) and \(h(3)=\sqrt4=2\), so the change is \(2-0=2\). The secant slope is \(2/(3-(-1))=2/4=\frac12\). The theorem predicts \(c\in(-1,3)\) such that
Multiplying both sides by \(2\sqrt{c+1}\) gives \(1=\sqrt{c+1}\), so \(c+1=1\) and \(c=0\), which is inside \((-1,3)\). Substitution checks \(h'(0)=1/(2\sqrt1)=\frac12\), matching the secant slope.
A Common Interval Mistake
A monotonicity claim applies only to pairs of points in the stated interval. A point outside that interval cannot be used to show that the function fails to be increasing there.
Worked Example: Use Points Inside the Interval
Consider \(q(x)=x^2\) on \((-1,2)\). A tempting but invalid argument is: “\(q(-1)=1\) and \(q(0)=0\), so moving from \(-1\) to \(0\) decreases the function; therefore \(q\) is not increasing on \((-1,2)\).” The problem is that \(-1\notin(-1,2)\), so this pair does not test the definition on that interval.
Choose two points that are both inside instead: \(-\frac12<0\), with \(-\frac12,0\in(-1,2)\). Their function values are
Since \(\frac14>0\), we have \(q(-\frac12)>q(0)\) even though \(-\frac12<0\). This valid pair shows that \(q\) is not increasing on \((-1,2)\). It is also consistent with the derivative: \(q'(x)=2x\), which is negative for \(x<0\) in the interval and therefore is not positive throughout it.
How to Use the Result Carefully
The derivative-sign theorem is a sufficient condition: a positive derivative throughout an interval guarantees strict increase, and a negative derivative guarantees strict decrease. The theorem does not say that these conditions are necessary. In particular, a derivative being zero at some points does not, by itself, settle whether a function is increasing or decreasing. The nonnegative-derivative conclusion guarantees only nondecrease directly from the Mean Value Theorem.
When writing a proof, be precise about the interval and the hypotheses. Choose arbitrary \(x_1<x_2\), verify the Mean Value Theorem conditions on \([x_1,x_2]\), and use the derivative sign at the resulting intermediate point. This short argument proves a statement about every pair, not merely about a few values or a visual impression.
Check Your Understanding
For each question, state the relevant interval and explain how the Mean Value Theorem supports the conclusion.
- If \(f\) is continuous on \([-2,4]\), differentiable on \((-2,4)\), and \(f'(x)<0\) throughout \((-2,4)\), what monotonicity conclusion follows on \([-2,4]\)?
- Why does the proof compare two arbitrary points rather than just evaluate the function at the interval’s endpoints?
- If \(f'(x)\geq 0\) throughout an interval and the Mean Value Theorem hypotheses hold, what conclusion follows? Is it necessarily strict increase from this result alone?
- For \(p(x)=x^2\) on \((-1,2)\), give a pair of points inside the interval that disproves increasing behavior.
- Must a function’s derivative exist at an endpoint to apply the Mean Value Theorem on a closed interval?
