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

1 Answer
Aug 11, 2015

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]}