The Mean Value Theorem

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is $y = x^3$ on $[0, 2]$ with its secant line, of slope 4. The blue line is the tangent at $x = c$ . Move $c$ until the tangent is parallel to the secant.

The curve y = x³, its secant line y = 4x on [0, 2], and its tangent at x = c. Near c ≈ 1.15 the tangent is parallel to the secant.

Is there a point in $(0, 2)$ where the tangent's slope equals 4?

Yes, near $c \approx 1.15$
Yes, at the midpoint $c = 1$

Key idea

Somewhere, the instantaneous rate equals the average rate.

Mean Value Theorem: if

f

is continuous on

[a, b]

and differentiable on

(a, b)

, there is a

c

in

(a, b)

with

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

.

Worked example

$f(x) = x^2$ on $[1, 5]$ . Find $c$ .

\dfrac{f(5) - f(1)}{5 - 1} = 6

Average rate: (25 − 1)/4.

2c = 6

f'(c) = 2c.

c = 3

Check: 3 is in (1, 5).

Your turn, guided

Your turn: $f(x) = x^2$ on $[2, 8]$ .

c = 5

(64 − 4)/6 = 10, so 2c = 10.

Practice ★

$f(x) = x^2$ on $[2, 5]$ . Find the $c$ given by the Mean Value Theorem.
$f(x) = x^2$ on $[3, 9]$ . Find the $c$ given by the Mean Value Theorem.
$f(x) = x^2$ on $[1, 7]$ . Find the $c$ given by the Mean Value Theorem.

Watch out!

Practice ★★

$f(x) = x^3$ on $[0, 4]$ . Find the $c$ in $(0, 4)$ given by the Mean Value Theorem.
$f(x) = x^3$ on $[0, 6]$ . Find the $c$ in $(0, 6)$ given by the Mean Value Theorem.

A car drove 276 miles in 4 hours. Must its speedometer have read exactly 69 mph at some moment?
1. Yes, by the Mean Value Theorem
2. Only if it drove at a constant speed
3. No: it may have gone faster and slower without hitting 69 exactly

Exit ticket

Three questions to check you have got it.

$f(x) = x^2$ on $[2, 5]$ . Find the $c$ given by the Mean Value Theorem.
$f(x) = x^2$ on $[3, 9]$ . Find the $c$ given by the Mean Value Theorem.
$f(x) = x^2$ on $[1, 7]$ . Find the $c$ given by the Mean Value Theorem.

Summary card

Mean Value Theorem

Check: continuous on $[a, b]$ , differentiable on $(a, b)$ .
Average rate $\dfrac{f(b) - f(a)}{b - a}$ .
Solve $f'(c) =$ that rate, with $c$ in $(a, b)$ .

Watch out: $c$ is not always the midpoint.

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