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).
Use the bound on f′
3f′(c) ≤ 3 · 2 = 6, so f(4) ≤ 5 + 6.
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.
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.
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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