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

1 Answer
Wataru
Sep 12, 2014

The Mean Value Theorem guarantees that there exists a number cc in (0,3)(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

Related questions
See all questions in Mean Value Theorem for Continuous Functions
Impact of this question
18576 views around the world
You can reuse this answer
Creative Commons License