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)#?

Question

Paisley Johnson · Accepted Answer

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

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#.

Ezekiel Alexander · Answer

#c=3 or c =-1# #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#

Paisley Johnson · Answer

undefined To determine all values of $ c $ that satisfy the Mean Value Theorem on the interval $[2,5]$ for $ f(x) = \frac{1}{x-1} $, we first find the average rate of change of $ f $ on the interval $[2,5]$. This is calculated as:

$$ f'(c) = \frac{f(5) - f(2)}{5 - 2} $$

$$ f'(c) = \frac{\frac{1}{5-1} - \frac{1}{2-1}}{5-2} $$

$$ f'(c) = \frac{\frac{1}{4} - 1}{3} $$

$$ f'(c) = \frac{-\frac{3}{4}}{3} $$

$$ f'(c) = -\frac{1}{4} $$

Now, we find the derivative of $ f(x) $:

$$ f'(x) = \frac{d}{dx}\left(\frac{1}{x-1}\right) = -\frac{1}{{(x-1)}^2} $$

To find $ c $, we set $ -\frac{1}{{(c-1)}^2} = -\frac{1}{4} $:

$$ -\frac{1}{{(c-1)}^2} = -\frac{1}{4} $$

$$ {(c-1)}^2 = 4 $$

$$ c-1 = \pm 2 $$

$$ c = 1 \pm 2 $$

Therefore, $ c $ can be $ 3 $ or $ -1 $. These are the values of $ c $ that satisfy the Mean Value Theorem on the interval $[2,5]$ for $ f(x) = \frac{1}{x-1} $.