Question

How do you find the number c that satisfies the conclusion of the Mean Value Theorem for `f( x) = e^(-2x)` on [0,3]?

Answer 1

Find slope `m` of secant of `f(x)` between `x = 0` and `x = 3`, then solve `f'(c) = m` to find:

`c = (ln(6) - ln(1-e^-6))/2 ~= 0.89712`

Explanation:

`f(0) = 1`, `f(3) = e^-6`, so the slope of the secant is

`(f(3)-f(0))/(3-0) = (e^-6-1)/3`

`f'(x) = -2e^(-2x)`

So `-2e^(-2c) = (e^-6-1)/3`

Divide both sides by `-2` to get:

`e^(-2c) = (1-e^-6)/6`

Take natural logs of both sides to get:

`-2c = ln(1-e^-6) - ln(6)`

Divide both sides by `-2` to get:

`c = (ln(6) - ln(1-e^-6))/2 ~= 0.89712`

As you might expect, this is fairly close to `ln(6) / 2 ~= 0.896`

graph{(y-e^(-2x))(y+x/3-1)(y+x/3-0.441) = 0 [-1.01, 3.99, -0.55, 1.95]}