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Sagot :
To solve the problem, we need to follow these steps:
1. Find the mean slope of [tex]\( f(x) = \frac{1}{x} \)[/tex] on the interval [tex]\([1, 11]\)[/tex]:
- First, calculate the value of [tex]\( f(x) \)[/tex] at the endpoints of the interval:
[tex]\[ f(1) = \frac{1}{1} = 1 \][/tex]
[tex]\[ f(11) = \frac{1}{11} \approx 0.09090909090909091 \][/tex]
- Next, find the difference in the function values at the endpoints:
[tex]\[ f(11) - f(1) \approx 0.09090909090909091 - 1 = -0.9090909090909091 \][/tex]
- The difference in [tex]\( x \)[/tex]-values at the endpoints is:
[tex]\[ 11 - 1 = 10 \][/tex]
- Therefore, the mean slope (or average rate of change) of the function on the interval [tex]\([1, 11]\)[/tex] is:
[tex]\[ \text{Mean Slope} = \frac{f(11) - f(1)}{11 - 1} \approx \frac{-0.9090909090909091}{10} = -0.09090909090909091 \][/tex]
2. Use the Mean Value Theorem to find [tex]\( c \)[/tex] such that [tex]\( f^{\prime}(c) \)[/tex] is equal to the mean slope:
- The derivative of [tex]\( f(x) = \frac{1}{x} \)[/tex] is:
[tex]\[ f^{\prime}(x) = -\frac{1}{x^2} \][/tex]
- According to the Mean Value Theorem, there exists a [tex]\( c \)[/tex] in the interval [tex]\((1, 11)\)[/tex] such that:
[tex]\[ f^{\prime}(c) = \text{Mean Slope} \][/tex]
- Set the derivative equal to the mean slope and solve for [tex]\( c \)[/tex]:
[tex]\[ -\frac{1}{c^2} = -0.09090909090909091 \][/tex]
- Simplify and solve for [tex]\( c \)[/tex]:
[tex]\[ \frac{1}{c^2} = 0.09090909090909091 \][/tex]
[tex]\[ c^2 = \frac{1}{0.09090909090909091} = 11 \][/tex]
[tex]\[ c = \sqrt{11} \text{ or } c = -\sqrt{11} \][/tex]
- Since [tex]\( c \)[/tex] must be in the interval [tex]\((1, 11)\)[/tex], we discard the negative value [tex]\( c = -\sqrt{11} \)[/tex].
[tex]\[ c = \sqrt{11} \approx 3.31662479035540 \][/tex]
Therefore, the [tex]\( c \)[/tex] that satisfies the Mean Value Theorem for the function [tex]\( f(x) = \frac{1}{x} \)[/tex] on the interval [tex]\((1, 11)\)[/tex] is approximately [tex]\( c \approx 3.31662479035540 \)[/tex].
1. Find the mean slope of [tex]\( f(x) = \frac{1}{x} \)[/tex] on the interval [tex]\([1, 11]\)[/tex]:
- First, calculate the value of [tex]\( f(x) \)[/tex] at the endpoints of the interval:
[tex]\[ f(1) = \frac{1}{1} = 1 \][/tex]
[tex]\[ f(11) = \frac{1}{11} \approx 0.09090909090909091 \][/tex]
- Next, find the difference in the function values at the endpoints:
[tex]\[ f(11) - f(1) \approx 0.09090909090909091 - 1 = -0.9090909090909091 \][/tex]
- The difference in [tex]\( x \)[/tex]-values at the endpoints is:
[tex]\[ 11 - 1 = 10 \][/tex]
- Therefore, the mean slope (or average rate of change) of the function on the interval [tex]\([1, 11]\)[/tex] is:
[tex]\[ \text{Mean Slope} = \frac{f(11) - f(1)}{11 - 1} \approx \frac{-0.9090909090909091}{10} = -0.09090909090909091 \][/tex]
2. Use the Mean Value Theorem to find [tex]\( c \)[/tex] such that [tex]\( f^{\prime}(c) \)[/tex] is equal to the mean slope:
- The derivative of [tex]\( f(x) = \frac{1}{x} \)[/tex] is:
[tex]\[ f^{\prime}(x) = -\frac{1}{x^2} \][/tex]
- According to the Mean Value Theorem, there exists a [tex]\( c \)[/tex] in the interval [tex]\((1, 11)\)[/tex] such that:
[tex]\[ f^{\prime}(c) = \text{Mean Slope} \][/tex]
- Set the derivative equal to the mean slope and solve for [tex]\( c \)[/tex]:
[tex]\[ -\frac{1}{c^2} = -0.09090909090909091 \][/tex]
- Simplify and solve for [tex]\( c \)[/tex]:
[tex]\[ \frac{1}{c^2} = 0.09090909090909091 \][/tex]
[tex]\[ c^2 = \frac{1}{0.09090909090909091} = 11 \][/tex]
[tex]\[ c = \sqrt{11} \text{ or } c = -\sqrt{11} \][/tex]
- Since [tex]\( c \)[/tex] must be in the interval [tex]\((1, 11)\)[/tex], we discard the negative value [tex]\( c = -\sqrt{11} \)[/tex].
[tex]\[ c = \sqrt{11} \approx 3.31662479035540 \][/tex]
Therefore, the [tex]\( c \)[/tex] that satisfies the Mean Value Theorem for the function [tex]\( f(x) = \frac{1}{x} \)[/tex] on the interval [tex]\((1, 11)\)[/tex] is approximately [tex]\( c \approx 3.31662479035540 \)[/tex].
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