Slope Unity in Mean Value Theorem: Average Meets Instant

The Mean Value Theorem says that a function continuous on a closed interval and differentiable inside it must, at some point, have an instantaneous rate of change equal to its average rate of change across the interval. Below: the statement, why each condition is needed, how to find that point, and a worked example.

What the Mean Value Theorem says

If a function \( f \) is continuous on the closed interval \( [a, b] \) and differentiable on the open interval \( (a, b) \), then there is at least one number \( c \) between \( a \) and \( b \) for which

\( f'(c) = \dfrac{f(b) – f(a)}{b – a} \)

The right-hand side is the average rate of change across the whole interval, which is the slope of the secant line joining the two endpoints. The left-hand side is the instantaneous rate of change at a single point. The theorem guarantees that somewhere inside the interval, the instant matches the average.

The everyday version: if you drive 120 miles in two hours, your average speed is 60 mph, and at some instant your speedometer must have read exactly 60. You cannot average 60 without ever being at 60.

Both conditions are doing work

Drop either hypothesis and the conclusion fails.

Continuity on the closed interval. Take \( f(x) = \frac{1}{x} \) on \( [-1, 1] \). The average rate of change is \( \frac{1 – (-1)}{1 – (-1)} = 1 \), but the derivative \( -\frac{1}{x^2} \) is negative everywhere it exists, so no \( c \) works. The function is not continuous at \( x = 0 \).

Differentiability on the open interval. Take \( f(x) = |x| \) on \( [-1, 1] \). The secant slope is 0, but the derivative is \( -1 \) to the left of zero and \( +1 \) to the right, and does not exist at zero. Again no \( c \).

Continuity is required at the endpoints; differentiability is not. A function can have a corner or a vertical tangent exactly at \( a \) or \( b \) and the theorem still holds.

How to find the point c

1. finding the average rate of change for the interval \( [a,b] \) using the formula \( f'(c) = \frac{f(b) – f(a)}{b – a} \)
2. finding the derivative of the function using rules of differentiation
3. setting the derivative equal to average rate and solving for \( x \), which will be the coordinate of the \( x \) point that we are looking for, and we call it \(c\)

• secant line passes through the first and the last point of the function, so by finding one of these points using \( (x,f(x)) \) for our function, and using point-slope form, (with the slope being average rate of change of the function in the interval), we get the equation for the secant line.

• to find the equation for the tangent line, we put the point that we found in step 3 above using the mean value theorem, into the original function to find the corresponding \( y \) value of that \( x \). using point-slope form for this \( (x,y) \) point with the slope being average rate of change of the function in the interval, we get the equation for the tangent line.

Worked example: f(x) = x³ on [1, 2]

Step 1 — average rate of change. \( f(1) = 1 \) and \( f(2) = 8 \), so \( \dfrac{f(2) – f(1)}{2 – 1} = \dfrac{8 – 1}{1} = 7 \).

Step 2 — the derivative. \( f'(x) = 3x^2 \).

Step 3 — set them equal and solve. \( 3c^2 = 7 \), so \( c^2 = \frac{7}{3} \) and \( c = \sqrt{7/3} \approx 1.53 \). The negative root is discarded because it falls outside \( [1, 2] \).

Step 4 — the secant line. It passes through \( (1, 1) \) with slope 7, so \( y – 1 = 7(x – 1) \), giving \( y = 7x – 6 \).

Step 5 — the tangent line at c. \( f(c) = c^3 \approx 3.56 \), and the slope there is also 7, so \( y – 3.56 = 7(x – 1.53) \), giving \( y \approx 7x – 7.1 \).

The two lines have the same slope, so they are parallel, and that is the whole geometric content of the theorem. The secant cuts the curve at both endpoints; the tangent touches it once, at \( x \approx 1.53 \), and sits below the secant.

Where students lose marks

• Solving \( f'(x) = 0 \) instead of \( f'(x) = \) the secant slope. That is the test for a critical point. The secant slope is only zero in the special case where \( f(a) = f(b) \).
• Keeping a root outside the interval. The theorem promises a \( c \) strictly between \( a \) and \( b \). A solution equal to an endpoint, or outside the interval, does not count.
• Checking differentiability at the endpoints. Only the open interval needs it.
• Assuming \( c \) is unique. The theorem says at least one. A wavy function can match its average slope at several points.

Rolle’s Theorem, the special case

When \( f(a) = f(b) \), the secant slope is zero and the Mean Value Theorem reduces to: there is a \( c \) in \( (a, b) \) with \( f'(c) = 0 \). That is Rolle’s Theorem, and it is the version used to prove that a differentiable function between two equal values must turn around somewhere.

Practice questions on the Mean Value Theorem

1. Find every value of \( c \) guaranteed by the Mean Value Theorem for \( f(x) = x^2 \) on \( [0, 4] \).
2. Does the Mean Value Theorem apply to \( f(x) = \sqrt{x} \) on \( [0, 9] \)? If so, find \( c \).
3. Explain why the theorem does not apply to \( f(x) = \frac{1}{x – 2} \) on \( [1, 3] \).
4. A car travels 150 miles in 2.5 hours. What does the Mean Value Theorem let you conclude about its speed at some instant?

Answers

1. Secant slope \( = \frac{16 – 0}{4} = 4 \); \( f'(x) = 2x \), so \( 2c = 4 \) and \( c = 2 \).
2. Yes. \( \sqrt{x} \) is continuous on \( [0, 9] \) and differentiable on \( (0, 9) \). Secant slope \( = \frac{3 – 0}{9} = \frac{1}{3} \); \( f'(x) = \frac{1}{2\sqrt{x}} \), so \( \frac{1}{2\sqrt{c}} = \frac{1}{3} \), giving \( \sqrt{c} = 1.5 \) and \( c = 2.25 \).
3. The function is undefined at \( x = 2 \), which lies inside the interval, so it is not continuous on \( [1, 3] \).
4. Average speed is 60 mph, so at some instant the car was traveling at exactly 60 mph.

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