Given the function #f(x) = 1 / (x-1)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [2,5] and find the c?

Question

Ezra Adams · Accepted Answer

#c=3 in (2,5)#.

The Mean Value Theorem states that, : If a function #f : [a,b] rarr RR# is (1) continuous on #[a,b]#, (2) derivable on #(a,b)#, then there exists at least one #c in (a,b)# such that, #(f(b)-f(a))/(b-a)=f'(c)# We see that, the given fun. #f# is a rational fun., and hence, it is cont. on [2,5] and differentiable on #(2,5)#. Therefore, by MVT, there must exist a #c in (2,5)#, s.t.#f'(c)=(f(5)-f(2))/(5-2)={1/(5-1)-1/(2-1)}/(3)=1/3(1/4-1)=1/3(-3/4)=-1/4..........(1)# And, #f(x)=1/(x-1) rArr f'(x)=-1/(x-1)^2........................(2)# Hence, by #(1) and (2)#, #-1/(c-1)^2=-1/4 rArr (c-1)^2=4# #rArr c-1=+-2 rArr c=1+-2 rArr c=3, or, -1# Since, #c=-1 !in (2,5),# we get, #c=3 in (2,5)#.

Ezra Adams · Answer

undefined To determine if the function $ f(x) = \frac{1}{x-1} $ satisfies the hypotheses of the Mean Value Theorem on the interval $[2,5]$, we need to check if it is continuous on $[2,5]$ and differentiable on $(2,5)$.

1. **Continuity**: The function $ f(x) = \frac{1}{x-1} $ is continuous on $[2,5]$ except at $x = 1$, as the denominator becomes zero there.

2. **Differentiability**: The function $ f(x) = \frac{1}{x-1} $ is differentiable on $(2,5)$ because it is the quotient of two differentiable functions, $1$ and $x-1$, and the denominator $x-1$ is not zero on $(2,5)$.

Since $ f(x) $ is continuous on $[2,5]$ and differentiable on $(2,5)$, it satisfies the hypotheses of the Mean Value Theorem.

Now, to find the value of $ c $ guaranteed by the Mean Value Theorem, we use the formula:

$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

where $ a = 2 $ and $ b = 5 $.

1. Calculate $ f(b) $ and $ f(a) $:
$$ f(5) = \frac{1}{5-1} = \frac{1}{4} $$
$$ f(2) = \frac{1}{2-1} = 1 $$

2. Calculate $ f'(c) $:
$$ f'(c) = \frac{\frac{1}{4} - 1}{5 - 2} = \frac{-3/4}{3} = -\frac{1}{4} $$

To find $ c $, solve the equation $ f'(c) = -\frac{1}{4} $:
$$ -\frac{1}{4} = \frac{1}{(c-1)^2} $$
$$ (c-1)^2 = -4 $$
$$ c - 1 = \pm 2i $$
$$ c = 1 \pm 2i $$

Since $ c $ must be a real number within the interval $(2,5)$, the only valid solution is $ c = 1 $. Therefore, $ c = 1 $ is the value guaranteed by the Mean Value Theorem.