Given the function #f(x) = (x)/(x+2)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,4] and find the c?

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Answer 1 · 2016-09-04T18:50:50

Is #f# continuous on #[1,4]#? is it differentiable on #(1,4)#? If "yes" to both then it satisfies the hypotheses.

To find #c#, solve #f'(x) = (f(4)-f(1))/(4-1)#. Keep only the solutions (if any) in #(1,4)#

The Mean Value Theorem has two hypotheses:

H1 : #f# is continuous on the closed interval #[a,b]#

H2 : #f# is differentiable on the open interval #(a,b)#.

In this question, #f(x) = x/(x+2)# , #a=1# and #b=4#.

This function is continuous on its domain, (all reals except #-2#), so it is continuous on #[1, 4]#

#f'(x)=2/(x+2)^2# which exists for all #x != -2#.
#-2# is not in #(1,4)#, so

#f# is differentiable on #(1, 4)#.

This function on this interval satisfies the hypotheses of the Mean Value Theorem.

Therefore, we know without any further work that

there is a #c# in #(a,b)# for which #f'(c) = (f(b)-f(a))/(b-a)#

We have also been asked to find the #c# we solve the equation and note the requirement #c in (1,4)#

We need to solve

#2/(x+2)^2 = (f(4)-f(1))/(4-1)#

#2/(x+2)^2 = (2/3-1/3)/(4-1) = (1/3)/3 = 1/9#

#(x+2)^2 = 18#

#x+2 = +- sqrt18#

#x = -2 +- 3sqrt2#

Note that the solutions are #2 +- sqrt 18# and #sqrt18# is between #4# and #5#, so #-2 + sqrt18# is in #(1,4)#.

Of the two solutions, only #-2 + 3 sqrt2# is in #(1,4)#, so there is only one #c# and it is

#c = -2 + 3 sqrt2#