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Math · Calculus I · Worked example

Bound a value with the Mean Value Theorem

Suppose f is differentiable, f(1) = 5 and f′(x) ≤ 2 for every x. How large can f(4) be?

Apply the theorem on [1, 4]

Some c in (1, 4) has f′(c) = (f(4) − f(1))/3, so the change in f is 3f′(c).

f⁡(4)−f⁡(1)=3f⁡′(c)

Use the bound on f′

3f′(c) ≤ 3 · 2 = 6, so f(4) ≤ 5 + 6.

5+3⋅2=11

Check that the bound is reached

f(x) = 2x + 3 has f(1) = 5 and f′(x) = 2, and f(4) = 11, so 11 is the best possible bound.

2⋅4+3=11

Result

f(4) ≤ 11, and f(x) = 2x + 3 reaches 11.

Your turn

A car’s odometer reads 120 km at 1:00 and 290 km at 3:00. Show that its speed was exactly 85 km/h at some moment.

Show the answer and explanation

The average speed is (290 − 120)/2 = 85 km/h, so by the Mean Value Theorem the speed equals 85 km/h at some time between 1:00 and 3:00.

Position is a differentiable function of time, and its derivative is the speed.

290−1202=85

Keep exploring

In Math, change the bound on f′ from 2 to 0.5: the rows then give f(4) ≤ 5 + 3 · 0.5 = 6.5.

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