How do you determine all values of c that satisfy the mean value theorem on the interval [2,5] for #f(x) = 1 / (x-1)#?

2 Answers
Apr 4, 2016

Answer:

#AA c \in [f(5), f(2)] = [1/4, 1]#, #EE y \in [2, 5]# such that #f(y) = c#.
Explanation below.

Explanation:

Since #f# is a quotient of continuous function :
#f(x) = (g(x))/(h(x)), g(x) = 1# and #h(x) = x-1#,
it is defined and continuous for all #x \in RR \backslash {1}# (#f(1)# is not defined because we would divide by zero).

Therefore, the mean value theorem is satisfied for all closed interval #[a, b] \subset RR \backslash {1}#.

Since #[2, 5] \subset RR \backslash {1}# and #[2, 5]# is closed, we have, by the mean value theorem, that #AA c \in [f(5), f(2)] = [1/4, 1]#, #EE y \in [2, 5]# such that #f(y) = c#.

Apr 4, 2016

Answer:

#c=3 or c =-1#

Explanation:

#f'(c)=(f(b)-f(a))/(b-a)#

#f'(c)=(f(5)-f(2))/(5-2)#

#-1/(c-1)^2 = (1/4 -1)/3 =(-3/4)/3 = -3/4 *1/3=-1/4#

#-1=-1/4 (c-1)^2#

#4 = c^2-2c+1#

#0=c^2-2c-3#

#(c-3)(c+1)=0#

#c=3 or c =-1#