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
Shura
Apr 4, 2016

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.

Bdub
Apr 4, 2016

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

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