From Input Uncertainty to Output Bounds
An input measurement is often known only to lie within an interval. If \(x\in[a,b]\), the corresponding output \(f(x)\) may also vary. To find its worst-case lower and upper bounds, we need to know how the function orders its values across the interval.
The sign of the derivative can provide that information. If \(f\) is increasing, larger inputs produce outputs that are at least as large, so the smallest and largest possible outputs come from the left and right endpoints. If \(f\) is decreasing, the endpoint roles are reversed.
These bounds are exact consequences of monotonicity, not linear estimates. The key is to check the derivative sign over the entire input interval before mapping its endpoints.
Monotonicity and the Derivative Sign
A function need not have a strictly positive derivative to be strictly increasing, nor a strictly negative derivative to be strictly decreasing. However, derivative signs give useful sufficient conditions: if \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f'(x)\ge 0\) throughout \((a,b)\), then \(f\) is nondecreasing on \([a,b]\). If \(f'(x)\le 0\) throughout \((a,b)\), then \(f\) is nonincreasing there.
In either case, the endpoint values give the exact lower and upper bounds. When \(f\) is continuous, the Intermediate Value Theorem also ensures that every value between those endpoint outputs occurs. Thus the output range is the closed interval between the two endpoint values.
Why the Derivative Sign Gives These Bounds
The justification follows from the Mean Value Theorem. Its conditions are that a function be continuous on a closed interval and differentiable on the interval’s interior. Under those conditions, the difference in function values equals a derivative at an interior point multiplied by the input difference.
Take any \(x_1,x_2\in[a,b]\) with \(x_1<x_2\).
There is some \(c\in(x_1,x_2)\) such that \(f(x_2)-f(x_1)=f'(c)(x_2-x_1)\).
Since \(x_2-x_1>0\), if \(f'(c)\ge0\), then \(f(x_2)-f(x_1)\ge0\). Hence \(f(x_1)\le f(x_2)\), which is precisely the nondecreasing condition.
For nondecreasing \(f\), choose \(x_1=a\) and \(x_2=x\), then \(x_1=x\) and \(x_2=b\). This gives \(f(a)\le f(x)\le f(b)\). If \(f'\le0\), the inequalities reverse.
This argument also explains why the derivative sign must hold across the interval: the point \(c\) supplied by the Mean Value Theorem could be anywhere between the two inputs being compared.
Worked Example: Increasing Function
Worked Example: Increasing Function
Suppose \(x\in[1.9,2.1]\) and \(f(x)=x^3+2x\). Find the exact output bounds.
First differentiate and check the sign on the whole interval:
Because \(x^2\ge0\), we have \(3x^2+2\ge2>0\) for every \(x\). Therefore \(f\) is increasing on \([1.9,2.1]\), so its smallest output is \(f(1.9)\) and its largest is \(f(2.1)\).
Write the endpoints as fractions to calculate exactly:
Thus the exact output range is \([10.659,13.461]\). To check the endpoint calculations, \(1.9^3+2(1.9)=6.859+3.8=10.659\), and \(2.1^3+2(2.1)=9.261+4.2=13.461\). Since the function is continuous, all intermediate output values are attained as well.
Worked Example: Decreasing Function
Worked Example: Decreasing Function
Suppose \(x\in[4.8,5.2]\) and \(g(x)=100/x\). Find the exact output bounds.
The function is defined and differentiable throughout this interval, which contains only positive inputs. Its derivative is
Therefore \(g\) is decreasing. The smallest output occurs at the largest input, \(5.2\), and the largest output occurs at the smallest input, \(4.8\):
Since \(\frac{250}{13}<\frac{125}{6}\), the output range is \(\left[\frac{250}{13},\frac{125}{6}\right]\). As a decimal description, these bounds are approximately \(19.23\) and \(20.83\), respectively, rounded to two decimal places. The exact fractions establish the bounds without rounding.
Check by substitution: \(100/(250/13)=100\cdot13/250=5.2\), and \(100/(125/6)=100\cdot6/125=4.8\). The smaller output corresponds to the larger input, as expected for a decreasing function.
When the Derivative Changes Sign
If the derivative changes sign inside the uncertainty interval, the function is not guaranteed to be monotone across the whole interval. In that case, do not simply compare the two endpoint outputs. Instead, divide the interval at points where the derivative changes sign, determine monotonicity on each piece, and compare the outputs at the resulting endpoints and turning points.
Worked Example: Splitting at a Sign Change
Let \(h(x)=x^2\) for \(x\in[-2,3]\). Find its output bounds.
The derivative is \(h'(x)=2x\). It is nonpositive on \([-2,0]\) and nonnegative on \([0,3]\), so \(h\) decreases on the first piece and increases on the second. Evaluate at the endpoints and at the split point:
On \([-2,0]\), the values decrease from \(4\) to \(0\); on \([0,3]\), they increase from \(0\) to \(9\). Hence the full output range is \([0,9]\). Checking the calculations directly gives \((-2)^2=4\), \(0^2=0\), and \(3^2=9\). The interval’s endpoint values alone would give \(4\) and \(9\), missing the true minimum at the interior point \(x=0\).
A Common Endpoint-Order Mistake
A frequent error is to assume that the lower input always gives the lower output. That is true for an increasing function, but false for a decreasing one. For example, for \(g(x)=100/x\) on \([4.8,5.2]\), writing the output interval as \([g(4.8),g(5.2)]\) would produce \(\left[\frac{125}{6},\frac{250}{13}\right]\). This is not an ordered interval, since \(\frac{125}{6}>\frac{250}{13}\).
The correction is to determine monotonicity first. Because \(g'(x)<0\), the function decreases, so the lower output is \(g(5.2)=250/13\) and the upper output is \(g(4.8)=125/6\). Always order the final bounds from smaller to larger, even when the input endpoints are listed from smaller to larger.
Practical Method
For an input known to lie in a closed interval \([a,b]\), use this sequence:
- Confirm that the function is defined on the entire interval.
- Find \(f'(x)\) and determine its sign throughout the interval.
- If the sign is nonnegative, use the increasing order \(f(a)\le f(x)\le f(b)\). If it is nonpositive, reverse the endpoint order.
- Evaluate both relevant endpoint values carefully and state the output interval from lower bound to upper bound.
- If the sign changes, split the input interval into pieces and compare the endpoint values on those pieces, including the sign-change points.
When the input endpoints represent measured quantities, preserve their units through the calculation: the output bounds have the units of \(f(x)\), not necessarily the units of \(x\). Keep exact values when convenient, and round only if a decimal report is needed.
Check Your Understanding
Answer each question by checking the derivative sign across the full interval.
- If \(f'(x)\ge0\) throughout \([a,b]\), which endpoint gives the lower bound for \(f(x)\)?
- For \(q(x)=12-3x\) on \([1,4]\), find the exact output bounds and state why the endpoint order is reversed.
- Why is it not enough to check that \(f'(a)\) and \(f'(b)\) have the same sign?
- For \(r(x)=x^2\) on \([-3,1]\), where must you split the interval to use monotonicity, and what are the output bounds?
- What continuity and differentiability conditions allow the Mean Value Theorem to justify a derivative-sign test on a closed interval?
