IDNLearn.com makes it easy to find precise answers to your specific questions. Get accurate and timely answers to your queries from our extensive network of experienced professionals.
Sagot :
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].
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.
