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

1 Answer
Sep 12, 2014

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#