When a Function Cannot Change Too Slowly
A bound on the derivative can restrict how much a function changes between two inputs. If the derivative has a fixed sign and its magnitude is never smaller than a positive number, then the function cannot change by an arbitrarily small amount over a fixed input distance. The Mean Value Theorem makes this idea precise.
This result has a useful consequence for equations. If a function must change by at least a fixed amount whenever its input changes, two distinct inputs cannot produce the same output. That can prove that an equation has at most one solution. It does not, by itself, prove that a solution exists.
The Lower-Bound Principle
Let \(I\) be an interval, and let \(m>0\). Suppose \(f\) is continuous on every closed segment contained in \(I\), differentiable in the interior of each such segment, and satisfies one of the following derivative bounds throughout the interior of \(I\): \(f'(t)\ge m\), or \(f'(t)\le -m\). The derivative therefore keeps a fixed sign and has magnitude at least \(m\).
The absolute-value inequality says that the function’s output change is at least \(m\) times the input change. The fixed-sign condition matters: a derivative whose magnitude is large but whose sign varies could produce changes in opposite directions on different parts of an interval. The theorem applies on intervals where one of the stated signed bounds holds.
Why the Bound Follows
Take any \(x,y\in I\) with \(x<y\). By the Mean Value Theorem, there is a point \(c\in(x,y)\) such that
If \(f'(t)\ge m\) throughout the interval, then \(f'(c)\ge m\). Since \(y-x>0\), multiplying the equation by \(y-x\) gives
In particular, \(f(y)-f(x)\) is positive, so its absolute value equals itself. Thus \(|f(y)-f(x)|\ge m|y-x|\).
If instead \(f'(t)\le -m\), then \(f'(c)\le -m\), and
This difference is negative, so \( |f(y)-f(x)|=-(f(y)-f(x))\ge m(y-x)\). These arguments cover \(x<y\). If \(y<x\), interchange the two points; both absolute values remain unchanged. When \(x=y\), both sides are zero. This proves the stated inequality for every pair of inputs.
Worked Example: A Positive Derivative Bound
Worked Example: Bounding a Change for a Quadratic
Let \(f(x)=x^2+3x\) on \([0,2]\). Find a guaranteed lower bound for the magnitude of the change from \(x=\frac12\) to \(x=\frac32\), and compare it with the exact change.
First, differentiate and bound the derivative on the interval:
so \(f'(x)\ge 3\). The theorem applies with \(m=3\). The input separation is
Therefore the magnitude of the function change is at least
Compute the function values to check the bound:
Thus the exact change is
The magnitude is \(5\), which is at least \(3\), as guaranteed. Substitution verifies the computed endpoint values, and their difference verifies the comparison.
Worked Example: A Negative Derivative Bound
Worked Example: Bounding a Decrease
Let \(g(x)=x^2-4x\) on \([0,1]\). Use a derivative bound to estimate the magnitude of the change from \(x=\frac14\) to \(x=\frac34\), then check the estimate directly.
The derivative is
For \(0\le x\le1\), we have \(2x\le2\), so \(g'(x)\le -2\). The theorem applies with \(m=2\) in its negative-derivative case. The input separation is
Consequently, the magnitude of the function change must satisfy
Now evaluate both outputs:
The change is
Its magnitude is \(\frac32\), which exceeds the guaranteed lower bound \(1\). The negative sign of the change agrees with the negative derivative bound.
Turning an Output Difference into an Input Bound
The same inequality can be rearranged. If \(m>0\) and \(|f(y)-f(x)|\ge m|y-x|\), division by \(m\) gives
This form is useful when the outputs are known or constrained. A small output difference forces the inputs to be close. In particular, if \(f(x)=f(y)\), the right-hand side is zero, so \(|y-x|=0\) and \(x=y\). Thus, on an interval where the theorem’s derivative bound holds, an equation \(f(x)=c\) can have at most one solution.
Worked Example: Proving a Solution Is Unique
Show that the equation \(x^3+x=2\) has exactly one real solution.
Define \(f(x)=x^3+x\). Then
for every real \(x\), because \(x^2\ge0\). The lower-bound theorem therefore applies on any interval in \(\mathbb R\), with \(m=1\).
First verify that \(x=1\) is a solution:
Now suppose \(x\) is any real solution. Then \(f(x)=2\), and also \(f(1)=2\). Applying the theorem to the two inputs \(x\) and \(1\) gives
The left-hand side equals \(|2-2|=0\). Hence \(0\ge|x-1|\). Since an absolute value is nonnegative, this forces \(|x-1|=0\), so \(x=1\). We have checked that \(1\) is a solution and proved that every solution must equal \(1\); therefore it is the unique real solution.
Worked Example: Constraining Inputs from Approximate Outputs
Worked Example: A Bound on the Distance Between Inputs
Let \(h(x)=5x+\sin x\) on \([-\frac{\pi}{2},\frac{\pi}{2}]\). Suppose \(x\) and \(y\) in this interval both satisfy \(|h(x)-7|\le0.01\) and \(|h(y)-7|\le0.01\). Bound the distance between \(x\) and \(y\).
Differentiate:
On the stated interval, \(\cos t\ge0\), so \(h'(t)\ge5\). The theorem applies with \(m=5\), giving
By the triangle inequality and the two assumed output bounds,
Combining the inequalities gives
Thus any two inputs whose outputs are within \(0.01\) of \(7\) must be no more than \(0.004\) apart. The bound follows from the stated derivative and output inequalities; it does not require either output to equal \(7\) exactly.
What the Result Does—and Does Not—Say
A lower derivative bound controls changes from below: inputs separated by a given distance must have outputs separated by at least the corresponding scaled distance. Conversely, outputs that are close can only come from inputs that are close. This is why the result is useful for uniqueness and for bounding the spread of possible solutions.
The conclusion is not an existence theorem. For example, proving that an equation has at most one solution does not show that its output level is ever reached. To establish that a solution exists, a separate argument is needed. In the cubic example, substitution supplied a solution; the derivative bound then established that there could not be another one.
The derivative condition must also apply throughout the interval between the inputs being compared. A bound verified only at the endpoints is not enough to use the Mean Value Theorem conclusion. Check both the hypotheses of that theorem and the sign and size of the derivative on the whole relevant interval.
For instance, in the first example \(f'(x)\ge3\) on \([0,2]\), and the change from \(\frac12\) to \(\frac32\) has magnitude \(5\). The incorrect claim \( |f(3/2)-f(1/2)|\le3\cdot1=3\) would assert \(5\le3\), which is false. The correct conclusion is \( |f(3/2)-f(1/2)|\ge3\).
A Reliable Application Process
Specify the interval containing the inputs being compared, and ensure the function is continuous on the relevant closed segment and differentiable in its interior.
Show either \(f'(t)\ge m>0\) everywhere on the interval or \(f'(t)\le-m<0\) everywhere there.
Use \(|f(y)-f(x)|\ge m|y-x|\) to bound a function change below, or rearrange it to bound input separation above.
A shared output proves uniqueness on the interval, but existence must be established separately.
Check Your Understanding
Use the derivative lower-bound principle and its hypotheses when answering these questions.
- If \(f'(t)\ge4\) on an interval, what lower bound follows for \(|f(y)-f(x)|\) in terms of \(|y-x|\)?
- If \(f'(t)\le-3\) on an interval and \(x<y\), what inequality does \(f(y)-f(x)\) satisfy?
- If \(|f(y)-f(x)|\ge 2|y-x|\) and \(|f(y)-f(x)|\le0.06\), what bound follows for \(|y-x|\)?
- Why does the derivative bound show that \(f(x)=c\) has at most one solution on the interval, but not necessarily that a solution exists?
- What goes wrong if a derivative bound has only been checked at the two endpoints rather than throughout the interval between them?
