Combining the Mean Value Theorem with Other Information
The Mean Value Theorem is most useful when it connects two kinds of information: the change in a function across an interval and the derivative at a point inside that interval. In applications, the theorem may help calculate an intermediate point, turn a derivative bound into a bound on function values, prove an inequality, or show that an equation has at most one solution.
These are not separate tricks. Each application uses the same central statement, together with conditions that must be checked. The derivative information is useful only on the interval where it is known, and the function must satisfy the theorem's hypotheses there.
The fraction on the right is the average rate of change of \(f\) across the interval. The theorem says that this average rate is attained as an instantaneous rate somewhere strictly between the endpoints. Before using the conclusion, verify continuity on the closed interval and differentiability on its interior.
Finding and Checking an Intermediate Point
When a problem asks for the point supplied by the theorem, first compute the average rate of change. Then solve the equation that sets the derivative equal to that rate. Any solution must lie in the open interval, and the function must meet the theorem's hypotheses.
Worked Example: Find a Mean Value Theorem Point
Find a point \(c\) for \(f(x)=x^3-3x\) on \([0,2]\). The function is a polynomial, so it is continuous on \([0,2]\) and differentiable on \((0,2)\). The theorem applies.
Compute the endpoint values:
The average rate of change is
Differentiate and set the derivative equal to \(1\):
Thus \(3c^2=4\), so \(c^2=\frac{4}{3}\). The solutions are \(c=\frac{2}{\sqrt{3}}\) and \(c=-\frac{2}{\sqrt{3}}\). Only the positive solution is in \((0,2)\), so the required point is \(c=\frac{2}{\sqrt{3}}\). It lies in the interval because \(0<\frac{2}{\sqrt{3}}<2\).
Check the derivative at that point:
The derivative equals the average rate of change, as required.
Turning Derivative Bounds into Difference Bounds
A bound on the derivative gives a bound on how much the function can change. The important conditions are that the derivative has a finite bound throughout the interior of the interval and that the Mean Value Theorem applies.
Here is the justification. By the Mean Value Theorem, there is a \(c\in(a,b)\) for which \(f(b)-f(a)=f'(c)(b-a)\). Taking absolute values and using both \(|f'(c)|\le M\) and \(b-a>0\) gives the claimed bound:
This is a complete deduction from the theorem and the derivative bound. In particular, the bound \(M\) must apply at the intermediate point the theorem supplies. A derivative bound that holds only at the endpoints, or only on part of the interval, is not enough.
Worked Example: Bound a Change in a Square-Root Function
Use the Mean Value Theorem to bound \(|\sqrt{4}-\sqrt{1}|\). Set \(f(x)=\sqrt{x}\) on \([1,4]\). This function is continuous on that interval and differentiable on its interior. For \(x\in(1,4)\),
because \(\sqrt{x}>1\) on \((1,4)\). The derivative-bound result with \(M=\frac12\) gives
The left-hand side is exactly \(|2-1|=1\), and \(1\le\frac32\), so the bound is confirmed. The intermediate point can also be found: the average rate is \(\frac{2-1}{4-1}=\frac13\), so \(f'(c)=\frac13\) means \(\frac{1}{2\sqrt{c}}=\frac13\). Hence \(2\sqrt{c}=3\), \(\sqrt{c}=\frac32\), and \(c=\frac94\), which lies in \((1,4)\). Checking gives \(f'(\frac94)=\frac{1}{2(3/2)}=\frac13\), the required average rate.
Using the Theorem to Prove an Inequality
A common strategy is to apply the theorem to a function whose endpoint values encode the expression in the desired inequality. The sign of the derivative at the resulting intermediate point can then determine the sign of the difference.
Worked Example: Prove \(\ln(1+x)\le x\)
We prove that \(\ln(1+x)\le x\) for every \(x>-1\). If \(x=0\), both sides are \(0\). Suppose first that \(x>0\). Apply the Mean Value Theorem to \(f(t)=\ln(1+t)\) on \([0,x]\). Since \(1+t>0\) throughout this interval, \(f\) is continuous on \([0,x]\) and differentiable on \((0,x)\). For some \(c\in(0,x)\),
Since \(c>0\), \(1+c>1\). As \(x>0\), division by \(1+c>1\) gives \(\frac{x}{1+c}<x\). Thus \(\ln(1+x)<x\) when \(x>0\).
Now suppose \(-1<x<0\). Apply the theorem on \([x,0]\), where \(1+t>0\), so the same continuity and differentiability conditions hold. For some \(c\in(x,0)\),
Equivalently, \(\ln(1+x)=\frac{x}{1+c}\). Because \(x<c<0\), we have \(0<1+c<1\). Dividing the negative number \(x\) by a positive number less than \(1\) makes it more negative, so \(\frac{x}{1+c}<x\). This proves the strict inequality for \(-1<x<0\) as well. Together with equality at \(x=0\), this establishes \(\ln(1+x)\le x\) for all \(x>-1\).
For a check, take \(x=1\). The argument gives \(\ln 2<1\); indeed, its intermediate point lies in \((0,1)\) and satisfies \(\ln 2=1/(1+c)\), which is less than \(1\).
Uniqueness from Strict Increase
The Mean Value Theorem can also show that two distinct inputs cannot produce the same function value. If \(f'(x)>0\) throughout an interval, then for any \(x_1<x_2\) in that interval the theorem gives a point \(c\in(x_1,x_2)\) such that \(f(x_2)-f(x_1)=f'(c)(x_2-x_1)>0\). Therefore \(f(x_2)>f(x_1)\): the function is strictly increasing. A strictly increasing function can take any particular value at most once.
Worked Example: Show an Equation Has Exactly One Solution
Show that \(x^3+x=1\) has exactly one solution in \([0,1]\). Define \(F(x)=x^3+x-1\), so solving the equation is equivalent to finding a zero of \(F\). Since \(F\) is a polynomial, it is continuous on \([0,1]\), and
The Intermediate Value Theorem therefore guarantees at least one zero in \((0,1)\). To prove there is at most one, differentiate:
For any \(x_1,x_2\in[0,1]\) with \(x_1<x_2\), the Mean Value Theorem applies on \([x_1,x_2]\). For some \(c\in(x_1,x_2)\),
since both factors are positive. Thus \(F\) is strictly increasing and has at most one zero. Combining existence with at-most-one gives exactly one solution in \([0,1]\). As a check on the uniqueness reasoning, if two distinct zeros \(x_1<x_2\) existed, the displayed equation would give \(0=F(x_2)-F(x_1)>0\), a contradiction.
A Common Pitfall: Using a Derivative Bound Where It Fails
A derivative-bound argument is only valid when the claimed bound holds throughout the open interval. For example, on \((0,1)\), the derivative of \(f(x)=\sqrt{x}\) is \(f'(x)=\frac{1}{2\sqrt{x}}\). It is not true that \(f'(x)\le\frac12\) there: at \(x=\frac14\), \(f'(\frac14)=1>\frac12\).
Consequently, the calculation \(|\sqrt{1}-\sqrt{0}|\le\frac12(1-0)\) is not justified by that derivative bound; its conclusion \(1\le\frac12\) is false. The error is not in the Mean Value Theorem, but in assuming a bound on the derivative that does not hold on the interval. On \([1,4]\), by contrast, \(f'(x)\le\frac12\) on the interior, so the earlier bound is valid there. Always check the derivative estimate on the whole open interval before using it.
Organizing a Mixed Application
When a problem combines several goals, separate the logical tasks. Establish the theorem's hypotheses first; then identify exactly what information is needed from the derivative. For an intermediate-point calculation, solve the average-rate equation and check the candidate. For an inequality, relate the relevant difference to a derivative value and establish its sign. For uniqueness, show strict monotonicity, and separately check existence if the question asks for exactly one solution.
Confirm continuity on the closed interval and differentiability on its interior before invoking the Mean Value Theorem.
For a point calculation, find the average rate of change. For a bound, identify a finite derivative bound. For uniqueness, determine the sign of the derivative.
The intermediate point lies inside the interval; a derivative bound must hold there; and a strictly increasing function has at most one solution to a given equation.
Substitute a proposed point into the derivative equation, check a numerical inequality directly when possible, or combine existence and uniqueness arguments explicitly.
Check Your Understanding
For each question, identify the relevant hypotheses as well as the calculation or conclusion.
- For \(f(x)=x^2\) on \([1,3]\), calculate the average rate of change and find a point \(c\in(1,3)\) where \(f'(c)\) equals it.
- If \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(|f'(x)|\le 4\) throughout \((a,b)\), what bound follows for \(|f(b)-f(a)|\)?
- In proving \(\ln(1+x)\le x\) for \(-1<x<0\), why does the intermediate point belong to \((x,0)\), and why does division by \(1+c\) make \(x\) more negative?
- What two separate facts establish that an equation has exactly one solution on an interval?
- Why is a derivative estimate that holds only at the endpoints insufficient for the derivative-bound result?
