Answer :
To determine whether the Mean Value Theorem (MVT) applies to the function [tex]\( f(x)=x^{-\frac{1}{3}} \)[/tex] on the interval [tex]\(\left[\frac{1}{27}, 27\right]\)[/tex], we need to verify two conditions:
1. The function must be continuous on the closed interval [tex]\([a, b]\)[/tex].
2. The function must be differentiable on the open interval [tex]\((a, b)\)[/tex].
Let's analyze [tex]\(f(x) = x^{-\frac{1}{3}}\)[/tex].
### Continuity
The function [tex]\( f(x) = x^{-\frac{1}{3}} \)[/tex] is a power function and is continuous for all [tex]\( x > 0 \)[/tex]. The interval [tex]\(\left[\frac{1}{27}, 27\right]\)[/tex] is entirely positive (since [tex]\( \frac{1}{27} > 0 \)[/tex]), so [tex]\( f(x) \)[/tex] is continuous on [tex]\(\left[\frac{1}{27}, 27\right]\)[/tex].
### Differentiability
Next, we need to check if [tex]\( f(x) \)[/tex] is differentiable on [tex]\((\frac{1}{27}, 27)\)[/tex]. The derivative of [tex]\( f(x) \)[/tex] is
[tex]\[ f'(x) = \left( x^{-\frac{1}{3}} \right)' = -\frac{1}{3} x^{-\frac{4}{3}}. \][/tex]
[tex]\( f'(x) \)[/tex] is defined for all [tex]\( x > 0 \)[/tex]. In our interval [tex]\( (\frac{1}{27}, 27) \)[/tex], all [tex]\( x \)[/tex] values are greater than [tex]\( 0 \)[/tex], so [tex]\( f(x) \)[/tex] is differentiable on [tex]\((\frac{1}{27}, 27)\)[/tex].
Since both conditions of the Mean Value Theorem are satisfied, we can say that:
### a. Determine whether the Mean Value Theorem applies to the given function on the interval [tex]\([a, b]\)[/tex].
The correct answer is:
C. Yes, because the function is continuous on the interval [tex]\(\left[\frac{1}{27}, 27\right]\)[/tex] and differentiable on the interval [tex]\(\left(\frac{1}{27}, 27\right)\)[/tex].
### b. Finding the point(s) guaranteed to exist by the Mean Value Theorem
The Mean Value Theorem states that if [tex]\( f \)[/tex] is continuous on [tex]\([a, b]\)[/tex] and differentiable on [tex]\( (a, b) \)[/tex], then there exists at least one [tex]\( c \in (a, b) \)[/tex] such that:
[tex]\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \][/tex]
Let's calculate the right side for our function:
[tex]\[ f(a) = \left( \frac{1}{27} \right)^{-\frac{1}{3}} = 27, \][/tex]
[tex]\[ f(b) = 27^{-\frac{1}{3}} = \frac{1}{3}, \][/tex]
[tex]\[ b - a = 27 - \frac{1}{27}. \][/tex]
Thus,
[tex]\[ \frac{f(b) - f(a)}{b - a} = \frac{\frac{1}{3} - 27}{27 - \frac{1}{27}}. \][/tex]
Simplify this expression:
[tex]\[ \frac{\frac{1}{3} - 27}{27 - \frac{1}{27}} = \frac{\frac{1}{3} - 27}{27 - \frac{1}{27}} = \frac{\frac{1}{3} - 27}{\frac{729 - 1}{27}} = \frac{\frac{1}{3} - 27}{\frac{728}{27}} = \frac{\frac{1}{3} - 27}{\frac{728}{27}} = \frac{\frac{1 - 81}{3}}{\frac{728}{27}} = \frac{\frac{-80}{3}}{\frac{728}{27}} = \frac{-80}{3} \times \frac{27}{728} = \frac{-80 \times 27}{3 \times 728} = \frac{-2160}{2184}.\][/tex]
Further simplification:
[tex]\[ \frac{-2160}{2184} = -0.989. \][/tex]
Next, set the derivative [tex]\( f'(x) = -\frac{1}{3} x^{-\frac{4}{3}} \)[/tex] equal to this value and solve for [tex]\( x \)[/tex]:
[tex]\[ -\frac{1}{3} x^{-\frac{4}{3}} = -0.989 \][/tex]
[tex]\[ x^{-\frac{4}{3}} = 2.967 \][/tex]
[tex]\[ x = \left( 2.967 \right)^{-\frac{3}{4}} \approx 0.461. \][/tex]
This point [tex]\( x \approx 0.461 \)[/tex] lies within [tex]\((\frac{1}{27}, 27)\)[/tex].
So, the point guaranteed to exist by the Mean Value Theorem is approximately [tex]\( c = 0.461 \)[/tex].
b. The correct choice is:
A. The point(s) guaranteed to exist is/are [tex]\( c = 0.461 \)[/tex].
1. The function must be continuous on the closed interval [tex]\([a, b]\)[/tex].
2. The function must be differentiable on the open interval [tex]\((a, b)\)[/tex].
Let's analyze [tex]\(f(x) = x^{-\frac{1}{3}}\)[/tex].
### Continuity
The function [tex]\( f(x) = x^{-\frac{1}{3}} \)[/tex] is a power function and is continuous for all [tex]\( x > 0 \)[/tex]. The interval [tex]\(\left[\frac{1}{27}, 27\right]\)[/tex] is entirely positive (since [tex]\( \frac{1}{27} > 0 \)[/tex]), so [tex]\( f(x) \)[/tex] is continuous on [tex]\(\left[\frac{1}{27}, 27\right]\)[/tex].
### Differentiability
Next, we need to check if [tex]\( f(x) \)[/tex] is differentiable on [tex]\((\frac{1}{27}, 27)\)[/tex]. The derivative of [tex]\( f(x) \)[/tex] is
[tex]\[ f'(x) = \left( x^{-\frac{1}{3}} \right)' = -\frac{1}{3} x^{-\frac{4}{3}}. \][/tex]
[tex]\( f'(x) \)[/tex] is defined for all [tex]\( x > 0 \)[/tex]. In our interval [tex]\( (\frac{1}{27}, 27) \)[/tex], all [tex]\( x \)[/tex] values are greater than [tex]\( 0 \)[/tex], so [tex]\( f(x) \)[/tex] is differentiable on [tex]\((\frac{1}{27}, 27)\)[/tex].
Since both conditions of the Mean Value Theorem are satisfied, we can say that:
### a. Determine whether the Mean Value Theorem applies to the given function on the interval [tex]\([a, b]\)[/tex].
The correct answer is:
C. Yes, because the function is continuous on the interval [tex]\(\left[\frac{1}{27}, 27\right]\)[/tex] and differentiable on the interval [tex]\(\left(\frac{1}{27}, 27\right)\)[/tex].
### b. Finding the point(s) guaranteed to exist by the Mean Value Theorem
The Mean Value Theorem states that if [tex]\( f \)[/tex] is continuous on [tex]\([a, b]\)[/tex] and differentiable on [tex]\( (a, b) \)[/tex], then there exists at least one [tex]\( c \in (a, b) \)[/tex] such that:
[tex]\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \][/tex]
Let's calculate the right side for our function:
[tex]\[ f(a) = \left( \frac{1}{27} \right)^{-\frac{1}{3}} = 27, \][/tex]
[tex]\[ f(b) = 27^{-\frac{1}{3}} = \frac{1}{3}, \][/tex]
[tex]\[ b - a = 27 - \frac{1}{27}. \][/tex]
Thus,
[tex]\[ \frac{f(b) - f(a)}{b - a} = \frac{\frac{1}{3} - 27}{27 - \frac{1}{27}}. \][/tex]
Simplify this expression:
[tex]\[ \frac{\frac{1}{3} - 27}{27 - \frac{1}{27}} = \frac{\frac{1}{3} - 27}{27 - \frac{1}{27}} = \frac{\frac{1}{3} - 27}{\frac{729 - 1}{27}} = \frac{\frac{1}{3} - 27}{\frac{728}{27}} = \frac{\frac{1}{3} - 27}{\frac{728}{27}} = \frac{\frac{1 - 81}{3}}{\frac{728}{27}} = \frac{\frac{-80}{3}}{\frac{728}{27}} = \frac{-80}{3} \times \frac{27}{728} = \frac{-80 \times 27}{3 \times 728} = \frac{-2160}{2184}.\][/tex]
Further simplification:
[tex]\[ \frac{-2160}{2184} = -0.989. \][/tex]
Next, set the derivative [tex]\( f'(x) = -\frac{1}{3} x^{-\frac{4}{3}} \)[/tex] equal to this value and solve for [tex]\( x \)[/tex]:
[tex]\[ -\frac{1}{3} x^{-\frac{4}{3}} = -0.989 \][/tex]
[tex]\[ x^{-\frac{4}{3}} = 2.967 \][/tex]
[tex]\[ x = \left( 2.967 \right)^{-\frac{3}{4}} \approx 0.461. \][/tex]
This point [tex]\( x \approx 0.461 \)[/tex] lies within [tex]\((\frac{1}{27}, 27)\)[/tex].
So, the point guaranteed to exist by the Mean Value Theorem is approximately [tex]\( c = 0.461 \)[/tex].
b. The correct choice is:
A. The point(s) guaranteed to exist is/are [tex]\( c = 0.461 \)[/tex].
