Question

How do I find the numbers `c` that satisfy the Mean Value Theorem for `f(x)=e^(-2x)` on the interval `[0,3]` ?

Answer 1

The Mean Value Theorem guarantees that there exists a number `c` in `(0,3)` such that `f'(c)={f(3)-f(0)}/{3-0}`. The actual value is
`c=ln sqrt{{6e^6}/{e^6-1}}approx0.9`.

Let us find such `c`.

By Chain Rule,
`f'(x)=-2e^{-2x}`

So, the left-hand side is
`f'(c)=-2e^{-2c}`

Let us find the right-hand side.
`{f(3)-f(0)}/{3-0}={e^{-6}-1}/3`
by multiplying the numerator and the denominator by `e^6`,
`={1-e^6}/{3e^6}`

By setting the left-hand side and the right-hand side equal to each other,
`-2e^{-2c}=={1-e^6}/{3e^6}`

By dividing by -2,
`e^{-2c}={e^6-1}/{3e^6}`

By taking the natural log,
`-2c=ln({e^6-1}/{6e^6})`

By dividing by -2,
`c=-1/2ln({e^6-1}/{6e^6})`

by the log property `rlnx=lnx^r`,
`=ln({e^6-1}/{6e^6})^{-1/2}`

by simplifying a bit further,
`=ln sqrt{{6e^6}/{e^6-1}}approx0.9`