Finding Mean Value Points for Rational and Trigonometric Functions

From the Secant Slope to a Mean Value Point

A function’s average rate of change over an interval is the slope of the line joining its endpoint values. The Mean Value Theorem guarantees that, under suitable conditions, this average rate is also the instantaneous rate of change at some interior point. For rational and trigonometric functions, finding that point usually means calculating the secant slope, differentiating, and solving an equation.

The task is not complete when an equation has been solved. A valid Mean Value Theorem point must lie strictly inside the interval, and the theorem’s hypotheses must hold on the whole interval. These checks are especially important for rational functions, which may be undefined at a denominator zero, and for trigonometric functions, whose derivatives depend on using radians.

The Theorem and the Calculation

The Mean Value Theorem: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, where $a<b$, then there is at least one $c\in(a,b)$ such that $$ f'(c)=\frac{f(b)-f(a)}{b-a}. $$ A value $c$ satisfying this equation is a Mean Value Theorem point for $f$ on $[a,b]$.

The fraction on the right is the secant slope: the total change in function value divided by the total change in input. The left side is the derivative at the interior point. Thus the calculation turns the theorem into an equation in $c$.

Here is the direct justification for that equation. If the tangent line at $x=c$ has the same slope as the secant line through $(a,f(a))$ and $(b,f(b))$, its slope must equal the secant’s slope. The tangent slope is $f'(c)$, so the equality is precisely the displayed equation. The hypotheses matter: the theorem guarantees at least one such interior point only when continuity and differentiability hold as stated. Solving the equation without checking those conditions does not establish a Mean Value Theorem point.

Check the interval.
Confirm that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. For a rational function, check that its denominator is nonzero throughout the closed interval.

Find the secant slope.
Calculate $\frac{f(b)-f(a)}{b-a}$, keeping the endpoint order consistent.

Set the derivative equal to that slope.
Compute $f'(x)$, then solve $f'(c)=\frac{f(b)-f(a)}{b-a}$.

Check the result.
Verify that the value lies in $(a,b)$ and substitute it into the derivative equation.

Worked Examples with Rational Functions

Worked Example: A Shifted Rational Function

Find a Mean Value Theorem point for $f(x)=\frac{x+1}{x+2}$ on $[0,2]$.

1. Check the hypotheses. The denominator is $x+2$. For every $x\in[0,2]$, $x+2\geq 2>0$. Thus the function is continuous on $[0,2]$ and differentiable on $(0,2)$.

2. Calculate the secant slope.

f(0)=\frac{1}{2},\qquad f(2)=\frac{3}{4},\qquad \frac{f(2)-f(0)}{2-0} = \frac{\frac{3}{4}-\frac{1}{2}}{2} = \frac{\frac{1}{4}}{2} = \frac{1}{8}.

3. Differentiate and solve. The quotient rule gives

f'(x) = \frac{(1)(x+2)-(x+1)(1)}{(x+2)^2} = \frac{x+2-x-1}{(x+2)^2} = \frac{1}{(x+2)^2}.

Set this equal to the secant slope and solve:

\frac{1}{(c+2)^2}=\frac{1}{8} \quad\Longrightarrow\quad (c+2)^2=8.

Since $c\in(0,2)$, $c+2>0$, so take the positive square root:

c+2=2\sqrt{2} \quad\Longrightarrow\quad c=2\sqrt{2}-2.

4. Check the interval and equation. Since $1<\sqrt{2}<2$, we have $0<2\sqrt{2}-2<2$. Also, $c+2=2\sqrt{2}$, so

f'(c)=\frac{1}{(2\sqrt{2})^2}=\frac{1}{8}.

This matches the secant slope, so $c=2\sqrt{2}-2$ is a Mean Value Theorem point.

Worked Example: A Reciprocal Function

Find a Mean Value Theorem point for $f(x)=\frac{1}{x}$ on $[1,3]$.

1. Check the hypotheses. The denominator $x$ is positive throughout $[1,3]$. The function is therefore continuous on the closed interval and differentiable on its interior.

2. Calculate the secant slope.

f(1)=1,\qquad f(3)=\frac{1}{3},\qquad \frac{f(3)-f(1)}{3-1} = \frac{\frac{1}{3}-1}{2} = \frac{-\frac{2}{3}}{2} = -\frac{1}{3}.

3. Differentiate and solve. Since $f'(x)=-\frac{1}{x^2}$, the required equation is

-\frac{1}{c^2}=-\frac{1}{3} \quad\Longrightarrow\quad c^2=3.

Because $c\in(1,3)$, $c$ is positive. The solution in that interval is $c=\sqrt{3}$.

4. Check the interval and equation. The inequalities $1<\sqrt{3}<3$ follow by squaring the positive quantities and using $1<3<9$. Substitution gives

f'(\sqrt{3}) = -\frac{1}{(\sqrt{3})^2} = -\frac{1}{3},

which equals the secant slope. Thus $c=\sqrt{3}$ is a Mean Value Theorem point.

Worked Example with a Trigonometric Function

Worked Example: The Sine Function

Find a Mean Value Theorem point for $f(x)=\sin x$ on $[0,\frac{\pi}{2}]$. Angles are measured in radians, as required for the derivative formula $\frac{d}{dx}\sin x=\cos x$.

1. Check the hypotheses. The sine function is continuous on $[0,\frac{\pi}{2}]$ and differentiable on its interior, so the Mean Value Theorem applies.

2. Calculate the secant slope.

f(0)=\sin 0=0,\qquad f\left(\frac{\pi}{2}\right)=\sin\left(\frac{\pi}{2}\right)=1,

\frac{f\left(\frac{\pi}{2}\right)-f(0)} {\frac{\pi}{2}-0} = \frac{1-0}{\frac{\pi}{2}} = \frac{2}{\pi}.

3. Differentiate and solve. Since $f'(x)=\cos x$, solve

\cos c=\frac{2}{\pi} \quad\Longrightarrow\quad c=\arccos\left(\frac{2}{\pi}\right).

4. Check the interval and equation. The inequalities $0<\frac{2}{\pi}<1$ hold because $\pi>2$. The principal value of $\arccos$ lies in $[0,\pi]$, and for an input strictly between 0 and 1 it lies in $(0,\frac{\pi}{2})$. Therefore $c$ is in the required open interval. By the definition of the inverse cosine,

f'(c) = \cos\left(\arccos\left(\frac{2}{\pi}\right)\right) = \frac{2}{\pi}.

This is the secant slope, so $c=\arccos(\frac{2}{\pi})$ is a Mean Value Theorem point. An inverse-trigonometric expression is an exact answer; a decimal approximation is not needed to identify or verify the point.

Common Pitfall: Confusing the Derivative Sign

For the sine example, a sign error can make the equation appear to have no solution. A common incorrect calculation is to write $f'(x)=-\cos x$, then conclude that $-\cos c=\frac{2}{\pi}$ is impossible because $\cos c\geq 0$ on $(0,\frac{\pi}{2})$. The error is the derivative: $\frac{d}{dx}\sin x=\cos x$, not $-\cos x$. The negative sign belongs to the derivative of cosine, $\frac{d}{dx}\cos x=-\sin x$. Using the correct derivative gives $\cos c=\frac{2}{\pi}$, which has the interior solution calculated above.

Another important check is that the candidate lies in the open interval, not merely in the closed interval. The endpoints $a$ and $b$ cannot serve as $c$ in the Mean Value Theorem conclusion. For rational functions, also check the denominator across the full closed interval before using the theorem; a denominator that is nonzero at the endpoints could still vanish somewhere between them.

What the Calculation Establishes

Each example follows the same structure even though the algebra differs. A rational derivative may lead to a square equation, while a trigonometric derivative may require an inverse-trigonometric function. In both cases, the secant slope supplies the target value for the derivative. The interval check then confirms that the resulting value has the location required by the theorem.

The theorem guarantees that at least one suitable point exists when its hypotheses hold. The calculation identifies a point by solving the derivative equation, and substitution verifies that the proposed value works. Keeping these roles separate is useful: the theorem supplies the existence guarantee, while the algebra finds and checks a point.

Check Your Understanding

For each question, state the secant slope or the relevant hypothesis before giving your answer.

For $f(x)=\frac{1}{x+1}$ on $[0,1]$, what equation must a Mean Value Theorem point satisfy?
For $f(x)=\cos x$ on $[0,\frac{\pi}{2}]$, calculate the secant slope and write the derivative equation for $c$.
Why must the denominator of a rational function be nonzero throughout the closed interval when checking continuity there?
For $f(x)=\sin x$, what derivative formula applies when $x$ is measured in radians?
After solving the derivative equation, what two checks are needed before accepting a candidate as a Mean Value Theorem point?

Finding Mean Value Points for Rational and Trigonometric Functions

What You'll Learn

From the Secant Slope to a Mean Value Point

The Theorem and the Calculation

Worked Examples with Rational Functions