One Guaranteed Input, Possibly Several
The average value of a continuous function on an interval is a horizontal level that the function reaches somewhere in that interval. The Mean Value Theorem for Integrals guarantees that such an input exists. It does not say that there is only one.
That distinction matters when a problem asks for all inputs where the function equals its average. A function can cross the average level several times, touch it, or—in the special case of a constant function—equal it everywhere. Finding one valid input is not the same as finding every solution.
The Average-Value Equation
For a continuous function \(f\) on \([a,b]\), its average value is the signed integral over the interval divided by the interval's length:
To find the inputs where the function equals its average value, first calculate this constant and then solve
Every solution must be checked against the interval being considered. If the question asks for inputs in \([a,b]\), include any endpoints that satisfy the equation. The theorem's guaranteed input, however, is in the open interval \((a,b)\).
The equation may have one solution, several solutions, or infinitely many. The integral gives the level to compare with the function; the function's shape and the equation determine how many times that level is reached.
Why Existence Does Not Mean Uniqueness
The Mean Value Theorem for Integrals says that if \(f\) is continuous on \([a,b]\), then there is at least one \(c\in(a,b)\) such that
Here is a short justification. A continuous function on a closed interval attains a minimum \(m\) and a maximum \(M\). Since \(m\leq f(x)\leq M\) throughout the interval, integrating and dividing by the positive length \(b-a\) gives
If \(f\) is nonconstant, its average lies strictly between its minimum and maximum. For example, at some point its value is greater than \(m\); by continuity, it remains greater than \(m\) on a small interval of positive length. Everywhere else it is at least \(m\), so the integral is strictly greater than \(m(b-a)\). Thus \(f_{\mathrm{avg}}>m\). The corresponding argument at a point where \(f(x)<M\) shows \(f_{\mathrm{avg}}<M\).
Choose points where \(f\) attains its minimum and maximum. By the Intermediate Value Theorem, between those points \(f\) takes every value strictly between \(m\) and \(M\), including \(f_{\mathrm{avg}}\). Since that average is strictly between the extreme values, a point attaining it is not an endpoint where an extreme occurs; in particular, a matching input can be found in \((a,b)\). If \(f\) is constant, its value equals its average at every point, so any \(c\in(a,b)\) works. This proves existence.
Nothing in the argument restricts how many times the function can take the average value. The Intermediate Value Theorem establishes that a value occurs; it does not say that it occurs only once. A function that rises and falls can return to the same horizontal level at different inputs.
Use the Equation to Find Every Solution
Worked Example: A Quadratic Has Two Matching Inputs
Find every \(x\in[-1,1]\) for which \(f(x)=x^2\) equals its average value on \([-1,1]\).
Step 1: Calculate the average value.
Step 2: Solve the average-value equation.
Step 3: Check the interval and the function values. Since \(0<1/\sqrt{3}<1\), both solutions lie in \([-1,1]\). Substitution gives
The equation \(x^2=1/3\) has exactly the two solutions shown, so both inputs—and no others—attain the average value. The graph is symmetric about the vertical axis, and the average level intersects its two sides at two points.
For an equation with multiple solutions, it is not enough to identify one crossing or to use the theorem as though it supplied a unique input. Solving the equation and checking its solutions against the interval is what establishes the complete answer.
Trigonometric Functions Can Reach the Average Repeatedly
Worked Example: Two Zeros at the Average Level
Find every \(x\in[0,1]\) for which \(f(x)=\cos(2\pi x)\) equals its average value on \([0,1]\).
Step 1: Calculate the average.
Step 2: Solve for where the function equals that average. The equation is
Cosine is zero at angles \(\pi/2+k\pi\), where \(k\) is an integer. Therefore
For \(0\leq x\leq1\), the integer values \(k=0\) and \(k=1\) give \(x=1/4\) and \(x=3/4\). Other integer values give inputs outside the interval.
Step 3: Check both answers.
Both values match \(f_{\mathrm{avg}}=0\), and the equation's full list of solutions on \([0,1]\) shows there are exactly two.
A Graph Can Reveal Several Crossings
A graph helps visualize the problem: draw the horizontal line \(y=f_{\mathrm{avg}}\) and identify every intersection with \(y=f(x)\) over the interval. But a graph is most reliable for suggesting the number and location of solutions; algebra or a sufficiently precise argument should confirm them.
Worked Example: Four Inputs Attain the Average
Find every \(x\in[0,1]\) for which \(f(x)=\cos(4\pi x)\) equals its average value on \([0,1]\).
Step 1: Find the average value.
The interval length is \(1\), so the factor \(1/(b-a)\) in the average-value formula is also \(1\).
Step 2: Solve \(f(x)=f_{\mathrm{avg}}\). The equation \(\cos(4\pi x)=0\) requires \(4\pi x=\pi/2+k\pi\), so
To find the values in \([0,1]\), use \(0\leq 1/8+k/4\leq1\). Subtracting \(1/8\) and multiplying by \(4\) gives \(-1/2\leq k\leq7/2\). The integer choices are \(k=0,1,2,3\), giving
Step 3: Verify each input. Substituting these four inputs makes \(4\pi x\) equal, respectively, to \(\pi/2\), \(3\pi/2\), \(5\pi/2\), and \(7\pi/2\). Cosine is zero at each of those angles, so all four function values equal the average \(0\). The integer restriction above shows there are no other solutions in the interval.
Read the Question's Interval Carefully
The examples show why a complete solution needs two checks: solve the equation and enforce the interval. In some problems, an endpoint may also equal the average. That endpoint is a solution on the closed interval, even though the Mean Value Theorem's guaranteed input is interior. Keep the domain of the question distinct from the theorem's conclusion.
The average-value equation can also have a single solution. For example, if \(f(x)=x\) on \([0,2]\), then
Solving \(x=1\) gives exactly one solution in \([0,2]\). The theorem guarantees at least one, while this particular equation establishes that there is exactly one. Uniqueness comes from the function and its equation, not from the theorem itself.
A Common Mistake: Taking Just One Root
For \(f(x)=x^2\) on \([-1,1]\), a tempting but incomplete calculation is
This drops the negative square root. The correction is to solve the equation with both possibilities:
Substitution confirms that both inputs give \(1/3\). In general, do not stop when you find a value that satisfies the average-value equation. Determine whether there are other solutions in the specified interval, and check each candidate.
Check Your Understanding
For each question, distinguish the average-value calculation from the step that establishes all solutions.
- For \(f(x)=x^2\) on \([-2,2]\), calculate the average value and solve \(f(x)=f_{\mathrm{avg}}\) on the interval.
- For \(f(x)=\sin(2\pi x)\) on \([0,1]\), calculate the average and find every input where the function equals it.
- What does the Mean Value Theorem for Integrals guarantee about the number and location of matching inputs?
- Why does finding one solution to \(f(x)=f_{\mathrm{avg}}\) not prove that it is the only solution?
- When solving the average-value equation, why must the candidate inputs be checked against the interval in the question?
