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?

Question

Emilia Aguilar · Accepted Answer

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#

Emily Ammerman · Answer

undefined To determine if the function satisfies the hypotheses of the Mean Value Theorem on the interval [1, 4], you need to check if the function is continuous on the closed interval [1, 4] and differentiable on the open interval (1, 4).

1. Check continuity: Evaluate the function at the endpoints of the interval, f(1) and f(4), to ensure it's continuous on the closed interval.
   - f(1) = 1/(1+2) = 1/3
   - f(4) = 4/(4+2) = 4/6 = 2/3

2. Check differentiability: Calculate the derivative of the function and check if it exists on the open interval (1, 4).
   - f'(x) = d/dx (x/(x+2))
          = [(x+2)(1) - (x)(1)] / (x+2)^2
          = 2/(x+2)^2

The function f(x) = x/(x+2) is continuous and differentiable on the interval (1, 4).

Now, to find the c that satisfies the Mean Value Theorem, apply the formula:
   - f'(c) = [f(4) - f(1)] / (4 - 1)

Substitute the values:
   - 2/(c+2)^2 = (2/3 - 1/3) / 3
   - 2/(c+2)^2 = 1/9

Solve for c:
   - (c+2)^2 = 18
   - c + 2 = ±√18
   - c = -2 ± √18

Since the interval is [1, 4], the value of c must be in the range (1, 4). Therefore, we discard the negative root. 
   - c = -2 + √18 ≈ 1.764

So, the value of c that satisfies the Mean Value Theorem on the interval [1, 4] is approximately 1.764.