Answer :
Let's analyze the given function [tex]\( f(x) = 4x^{-\frac{1}{3}} \)[/tex] on the interval [tex]\(\left[\frac{1}{8}, 8\right]\)[/tex] to determine whether the Mean Value Theorem (MVT) applies.
### Step-by-Step Solution
#### Part (a): Applying Mean Value Theorem (MVT)
1. Check Continuity:
The first condition for MVT to hold is that the function must be continuous on the closed interval [tex]\([a, b] = \left[\frac{1}{8}, 8\right]\)[/tex].
For [tex]\( f(x) = 4x^{-\frac{1}{3}} \)[/tex]:
- The function is defined for all [tex]\( x \)[/tex] in the interval [tex]\(\left[\frac{1}{8}, 8\right]\)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\(\frac{1}{8}\)[/tex] from the right and [tex]\(8\)[/tex] from the left, the function does not approach infinity or any undefined value.
Hence, it can be inferred that [tex]\( f(x) \)[/tex] is continuous on [tex]\(\left[\frac{1}{8}, 8\right]\)[/tex].
2. Check Differentiability:
The second condition for MVT to apply is that the function must be differentiable on the open interval [tex]\((a, b) = \left(\frac{1}{8}, 8\right]\)[/tex].
Take the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 4x^{-\frac{1}{3}} \][/tex]
[tex]\[ f'(x) = \frac{d}{dx} \left( 4x^{-\frac{1}{3}} \right) = 4 \left( -\frac{1}{3} \right) x^{-\frac{1}{3} - 1} = -\frac{4}{3} x^{-\frac{4}{3}} \][/tex]
- The derivative exists for all [tex]\( x \)[/tex] in the interval [tex]\(\left(\frac{1}{8}, 8\right)\)[/tex].
Therefore, [tex]\( f(x) \)[/tex] is differentiable on [tex]\(\left(\frac{1}{8}, 8\right]\)[/tex].
Conclusion:
Since [tex]\( f(x) = 4x^{-\frac{1}{3}} \)[/tex] is continuous on [tex]\(\left[\frac{1}{8}, 8\right]\)[/tex] and differentiable on [tex]\(\left(\frac{1}{8}, 8\right]\)[/tex], the Mean Value Theorem applies.
Thus, the correct answer is:
D. Yes, because the function is continuous on the interval [tex]\(\left[\frac{1}{8}, 8\right]\)[/tex] and differentiable on the interval [tex]\(\left(\frac{1}{8}, 8\right]\)[/tex].
#### Part (b): Find the Point(s) Guaranteed by the MVT
If the MVT applies, it guarantees at least one point [tex]\( c \)[/tex] in the interval [tex]\(\left(\frac{1}{8}, 8\right)\)[/tex] such that:
[tex]\[ f'(c) = \frac{f(b) - f(a)}{b - a} \][/tex]
First, calculate [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex]:
[tex]\[ f\left(\frac{1}{8}\right) = 4 \left(\frac{1}{8}\right)^{-\frac{1}{3}} = 4 \cdot 2 = 8 \][/tex]
[tex]\[ f(8) = 4 \cdot 8^{-\frac{1}{3}} = 4 \cdot \frac{1}{2} = 2 \][/tex]
Now, calculate the slope:
[tex]\[ \frac{f(b) - f(a)}{b - a} = \frac{2 - 8}{8 - \frac{1}{8}} = \frac{-6}{8 - 0.125} = \frac{-6}{7.875} = -\frac{24}{31.5} = -\frac{8}{10.5} = -\frac{8}{10.5} = -\frac{16}{21} \][/tex]
Next, solve for [tex]\( c \)[/tex] such that [tex]\( f'(c) = -\frac{16}{21} \)[/tex]:
[tex]\[ f'(x) = -\frac{4}{3} x^{-\frac{4}{3}} = -\frac{16}{21} \][/tex]
[tex]\[ -\frac{4}{3} x^{-\frac{4}{3}} = -\frac{16}{21} \][/tex]
[tex]\[ \frac{4}{3} x^{-\frac{4}{3}} = \frac{16}{21} \][/tex]
[tex]\[ x^{-\frac{4}{3}} = \frac{16}{21} \cdot \frac{3}{4} = \frac{12}{21} = \frac{4}{7} \][/tex]
[tex]\[ x^{-\frac{4}{3}} = \frac{4}{7} \][/tex]
[tex]\[ x^{4/3} = \frac{7}{4} \][/tex]
[tex]\[ x = \left( \frac{7}{4} \right)^{3/4} \][/tex]
No valid solution exists within the interval [tex]\(\left(\frac{1}{8}, 8\right)\)[/tex].
Therefore, the correct answer is:
B. The Mean Value Theorem does not apply to [tex]\(f(x)\)[/tex].
### Step-by-Step Solution
#### Part (a): Applying Mean Value Theorem (MVT)
1. Check Continuity:
The first condition for MVT to hold is that the function must be continuous on the closed interval [tex]\([a, b] = \left[\frac{1}{8}, 8\right]\)[/tex].
For [tex]\( f(x) = 4x^{-\frac{1}{3}} \)[/tex]:
- The function is defined for all [tex]\( x \)[/tex] in the interval [tex]\(\left[\frac{1}{8}, 8\right]\)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\(\frac{1}{8}\)[/tex] from the right and [tex]\(8\)[/tex] from the left, the function does not approach infinity or any undefined value.
Hence, it can be inferred that [tex]\( f(x) \)[/tex] is continuous on [tex]\(\left[\frac{1}{8}, 8\right]\)[/tex].
2. Check Differentiability:
The second condition for MVT to apply is that the function must be differentiable on the open interval [tex]\((a, b) = \left(\frac{1}{8}, 8\right]\)[/tex].
Take the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 4x^{-\frac{1}{3}} \][/tex]
[tex]\[ f'(x) = \frac{d}{dx} \left( 4x^{-\frac{1}{3}} \right) = 4 \left( -\frac{1}{3} \right) x^{-\frac{1}{3} - 1} = -\frac{4}{3} x^{-\frac{4}{3}} \][/tex]
- The derivative exists for all [tex]\( x \)[/tex] in the interval [tex]\(\left(\frac{1}{8}, 8\right)\)[/tex].
Therefore, [tex]\( f(x) \)[/tex] is differentiable on [tex]\(\left(\frac{1}{8}, 8\right]\)[/tex].
Conclusion:
Since [tex]\( f(x) = 4x^{-\frac{1}{3}} \)[/tex] is continuous on [tex]\(\left[\frac{1}{8}, 8\right]\)[/tex] and differentiable on [tex]\(\left(\frac{1}{8}, 8\right]\)[/tex], the Mean Value Theorem applies.
Thus, the correct answer is:
D. Yes, because the function is continuous on the interval [tex]\(\left[\frac{1}{8}, 8\right]\)[/tex] and differentiable on the interval [tex]\(\left(\frac{1}{8}, 8\right]\)[/tex].
#### Part (b): Find the Point(s) Guaranteed by the MVT
If the MVT applies, it guarantees at least one point [tex]\( c \)[/tex] in the interval [tex]\(\left(\frac{1}{8}, 8\right)\)[/tex] such that:
[tex]\[ f'(c) = \frac{f(b) - f(a)}{b - a} \][/tex]
First, calculate [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex]:
[tex]\[ f\left(\frac{1}{8}\right) = 4 \left(\frac{1}{8}\right)^{-\frac{1}{3}} = 4 \cdot 2 = 8 \][/tex]
[tex]\[ f(8) = 4 \cdot 8^{-\frac{1}{3}} = 4 \cdot \frac{1}{2} = 2 \][/tex]
Now, calculate the slope:
[tex]\[ \frac{f(b) - f(a)}{b - a} = \frac{2 - 8}{8 - \frac{1}{8}} = \frac{-6}{8 - 0.125} = \frac{-6}{7.875} = -\frac{24}{31.5} = -\frac{8}{10.5} = -\frac{8}{10.5} = -\frac{16}{21} \][/tex]
Next, solve for [tex]\( c \)[/tex] such that [tex]\( f'(c) = -\frac{16}{21} \)[/tex]:
[tex]\[ f'(x) = -\frac{4}{3} x^{-\frac{4}{3}} = -\frac{16}{21} \][/tex]
[tex]\[ -\frac{4}{3} x^{-\frac{4}{3}} = -\frac{16}{21} \][/tex]
[tex]\[ \frac{4}{3} x^{-\frac{4}{3}} = \frac{16}{21} \][/tex]
[tex]\[ x^{-\frac{4}{3}} = \frac{16}{21} \cdot \frac{3}{4} = \frac{12}{21} = \frac{4}{7} \][/tex]
[tex]\[ x^{-\frac{4}{3}} = \frac{4}{7} \][/tex]
[tex]\[ x^{4/3} = \frac{7}{4} \][/tex]
[tex]\[ x = \left( \frac{7}{4} \right)^{3/4} \][/tex]
No valid solution exists within the interval [tex]\(\left(\frac{1}{8}, 8\right)\)[/tex].
Therefore, the correct answer is:
B. The Mean Value Theorem does not apply to [tex]\(f(x)\)[/tex].
