Why Compare Derivatives?
Two functions can have different values and different shapes, yet their rates of change can be compared throughout an interval. The Mean Value Theorem turns that rate comparison into a precise comparison of how much the functions change from a shared reference point.
The key is to compare increments, such as \(f(x)-f(c)\), rather than immediately comparing \(f(x)\) and \(g(x)\). The increments measure how far each function has moved from its own value at \(c\). If the functions also have the same value at \(c\), then comparing their increments gives a comparison of their values.
The Comparison Theorem
The functions need to satisfy the same regularity conditions as in the Mean Value Theorem on the interval being considered. The derivative comparison must hold throughout the interval between the reference point and the point of comparison, not just at its endpoints or at one selected point.
- If \(x>c\), \(f(x)-f(c)\leq g(x)-g(c)\).
- If \(x<c\), \(f(x)-f(c)\geq g(x)-g(c)\).
- If \(x=c\), the two increments are both zero.
The reversal for \(x<c\) is important. In the Mean Value Theorem, a difference in function values is a derivative value multiplied by the displacement between the inputs. That displacement is positive to the right of \(c\) and negative to the left.
Why the Comparison Holds
Define \(h=f-g\). The derivative assumption says \(h'(t)=f'(t)-g'(t)\leq 0\) throughout the relevant open segment. For any \(x\ne c\), apply the Mean Value Theorem to \(h\) on the closed interval whose endpoints are \(c\) and \(x\). It gives a point \(\xi\) strictly between those endpoints such that the change in \(h\) equals its derivative at \(\xi\) times the displacement.
Since \(h'(\xi)\leq 0\), the product is nonpositive when \(x>c\), and nonnegative when \(x<c\). Replacing \(h\) by \(f-g\) yields exactly the two increment comparisons in the theorem. If \(x=c\), both increments equal zero, so the result holds there as well.
Worked Examples
Worked Example: Comparing Polynomial Increments to the Right
Let \(f(t)=t^2-2t\) and \(g(t)=t^2\) on \([1,3]\), with reference point \(c=1\). First differentiate:
For every \(t\in(1,3)\), \(f'(t)\leq g'(t)\), because \(2t-2\leq 2t\). The theorem therefore gives, for \(x\in[1,3]\), \(f(x)-f(1)\leq g(x)-g(1)\). At \(x=3\), calculate each increment:
The comparison is \(4\leq 8\), as predicted. Directly, the difference of the two increments is \(8-4=4\geq0\), confirming the result for this point.
Worked Example: A Different Pair of Derivatives
Let \(f(t)=t\) and \(g(t)=t^2\) on \([1,3]\), with \(c=1\). Their derivatives are \(f'(t)=1\) and \(g'(t)=2t\). For every \(t\in(1,3)\), \(1\leq2t\), so the theorem applies. At \(x=2\), it predicts \(f(2)-f(1)\leq g(2)-g(1)\). Evaluate the increments:
Thus \(1\leq3\). The derivative comparison and the computed increments agree. In this example, \(f(1)=g(1)=1\), so the increment comparison also gives \(f(2)\leq g(2)\), namely \(2\leq4\).
Worked Example: Comparing Increments to the Left
Let \(f(t)=t^2\) and \(g(t)=3t^2\) on \([0,2]\), and choose \(c=2\). Their derivatives are \(f'(t)=2t\) and \(g'(t)=6t\). On \((0,2)\), \(2t\leq6t\). Since \(x=1\) is to the left of \(c\), the theorem gives the reversed increment comparison \(f(1)-f(2)\geq g(1)-g(2)\). Compute:
Indeed, \(-3\geq-9\). The increments are negative because each function has a smaller value at \(1\) than at \(2\); the function with the smaller derivative has the less negative change when moving left. The calculated difference is \((-3)-(-9)=6\geq0\), checking the direction.
Increments Are Not Automatically Function Values
The theorem compares \(f(x)-f(c)\) with \(g(x)-g(c)\). It does not, by itself, say that \(f(x)\leq g(x)\). Rearranging the right-hand comparison gives
Thus the difference \(f-g\) at \(x\) is no larger than its difference at the reference point. To conclude \(f(x)\leq g(x)\), one also needs \(f(c)\leq g(c)\). In particular, when \(f(c)=g(c)\), the increment comparison for \(x\geq c\) implies \(f(x)\leq g(x)\).
This distinction matters whenever the starting values differ: derivative information controls how the gap between the functions changes, but it does not determine the gap’s initial value.
A Common Direction Error
A frequent mistake is to use the right-hand conclusion without checking whether \(x\) is to the right of the reference point. For the third example, an incorrect calculation might say, “Since \(f'(t)\leq g'(t)\), \(f(1)-f(2)\leq g(1)-g(2)\),” and then report \(-3\leq-9\). That inequality is false.
The correction is to use the Mean Value Theorem with the actual displacement. Here \(x-c=1-2=-1\), so \(h'(\xi)(x-c)\) is nonnegative when \(h'(\xi)\leq0\). Therefore the correct comparison is \(f(1)-f(2)\geq g(1)-g(2)\), which the values \(-3\) and \(-9\) confirm.
When the Result Applies
To use this comparison reliably, check each of the following:
- Both functions are continuous on the closed segment between \(c\) and \(x\).
- Both functions are differentiable in the interior of that segment.
- The derivative ordering \(f'(t)\leq g'(t)\) holds at every interior point of that segment.
- The direction of the final inequality matches the sign of \(x-c\).
The comparison need not hold outside the segment being considered. For a particular \(x\), it is enough to know the derivative ordering between \(c\) and \(x\). Conversely, knowing the derivative ordering at just one point does not meet the condition needed for the Mean Value Theorem argument.
The central conclusion is a comparison of changes from a common reference point. The Mean Value Theorem supplies the link between derivative ordering and those changes; the location of \(x\) relative to \(c\) determines the direction of the inequality.
Check Your Understanding
Use the derivative comparison theorem and pay attention to the reference point and the direction of each increment.
- If \(f'(t)\leq g'(t)\) between \(c\) and \(x\), what increment inequality follows when \(x>c\)?
- What is the corresponding inequality when \(x<c\), and why does its direction change?
- Let \(f(c)=5\) and \(g(c)=2\). Does a comparison of increments alone establish \(f(x)\leq g(x)\)? Explain.
- Which continuity and differentiability conditions are required to apply the Mean Value Theorem to the difference of two functions?
- Why is a derivative ordering at only one point not enough for this comparison theorem?
