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Sagot :
To solve this problem, let's work through it in a step-by-step manner.
1. Define the function and the interval:
The function given is [tex]\( f(x) = 6\sqrt{x} + 10 \)[/tex]. The interval we are interested in is [tex]\([1, 7]\)[/tex].
2. Compute [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex]:
We need to evaluate the function at the endpoints of the interval.
- For [tex]\( a = 1 \)[/tex]:
[tex]\[ f(1) = 6\sqrt{1} + 10 = 6 \cdot 1 + 10 = 16 \][/tex]
- For [tex]\( b = 7 \)[/tex]:
[tex]\[ f(7) = 6\sqrt{7} + 10 \][/tex]
3. Compute the mean slope of the function on the interval [tex]\([1, 7]\)[/tex]:
The mean slope is given by:
[tex]\[ \frac{f(b) - f(a)}{b - a} \][/tex]
Substituting the values, we get:
[tex]\[ \frac{(6\sqrt{7} + 10) - 16}{7 - 1} = \frac{6\sqrt{7} + 10 - 16}{6} = \frac{6\sqrt{7} - 6}{6} = \sqrt{7} - 1 \][/tex]
4. Find the derivative of [tex]\( f(x) \)[/tex]:
The derivative of the function [tex]\( f(x) = 6\sqrt{x} + 10 \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx} \left( 6\sqrt{x} + 10 \right) = 6 \cdot \frac{1}{2} \cdot x^{-1/2} = \frac{3}{\sqrt{x}} \][/tex]
5. Set the derivative equal to the mean slope to find [tex]\( c \)[/tex]:
According to the Mean Value Theorem, there exists some [tex]\( c \)[/tex] in the open interval [tex]\( (1, 7) \)[/tex] such that [tex]\( f'(c) \)[/tex] is equal to the mean slope. So, we set:
[tex]\[ f'(c) = \sqrt{7} - 1 \][/tex]
[tex]\[ \frac{3}{\sqrt{c}} = \sqrt{7} - 1 \][/tex]
6. Solve for [tex]\( c \)[/tex]:
Rearrange the equation to solve for [tex]\( c \)[/tex]:
[tex]\[ 3 = (\sqrt{7} - 1)\sqrt{c} \][/tex]
[tex]\[ \sqrt{c} = \frac{3}{\sqrt{7} - 1} \][/tex]
Squaring both sides to solve for [tex]\( c \)[/tex]:
[tex]\[ c = \left( \frac{3}{\sqrt{7} - 1} \right)^2 \][/tex]
7. Simplify the expression for [tex]\( c \)[/tex]:
Simplify the right side of the equation:
[tex]\[ c = \left( \frac{3}{\sqrt{7} - 1} \right)^2 = \frac{9}{(\sqrt{7} - 1)^2} \][/tex]
Therefore, the value of [tex]\( c \)[/tex] in the interval [tex]\( (1, 7) \)[/tex] that satisfies the Mean Value Theorem is:
[tex]\[ c = \frac{9}{(\sqrt{7} - 1)^2} \][/tex]
Thus, we have found that the [tex]\( c \)[/tex] such that [tex]\( f'(c) = \sqrt{7} - 1 \)[/tex] within the interval [tex]\( (1, 7) \)[/tex] is [tex]\( \frac{9}{(\sqrt{7} - 1)^2} \)[/tex].
1. Define the function and the interval:
The function given is [tex]\( f(x) = 6\sqrt{x} + 10 \)[/tex]. The interval we are interested in is [tex]\([1, 7]\)[/tex].
2. Compute [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex]:
We need to evaluate the function at the endpoints of the interval.
- For [tex]\( a = 1 \)[/tex]:
[tex]\[ f(1) = 6\sqrt{1} + 10 = 6 \cdot 1 + 10 = 16 \][/tex]
- For [tex]\( b = 7 \)[/tex]:
[tex]\[ f(7) = 6\sqrt{7} + 10 \][/tex]
3. Compute the mean slope of the function on the interval [tex]\([1, 7]\)[/tex]:
The mean slope is given by:
[tex]\[ \frac{f(b) - f(a)}{b - a} \][/tex]
Substituting the values, we get:
[tex]\[ \frac{(6\sqrt{7} + 10) - 16}{7 - 1} = \frac{6\sqrt{7} + 10 - 16}{6} = \frac{6\sqrt{7} - 6}{6} = \sqrt{7} - 1 \][/tex]
4. Find the derivative of [tex]\( f(x) \)[/tex]:
The derivative of the function [tex]\( f(x) = 6\sqrt{x} + 10 \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx} \left( 6\sqrt{x} + 10 \right) = 6 \cdot \frac{1}{2} \cdot x^{-1/2} = \frac{3}{\sqrt{x}} \][/tex]
5. Set the derivative equal to the mean slope to find [tex]\( c \)[/tex]:
According to the Mean Value Theorem, there exists some [tex]\( c \)[/tex] in the open interval [tex]\( (1, 7) \)[/tex] such that [tex]\( f'(c) \)[/tex] is equal to the mean slope. So, we set:
[tex]\[ f'(c) = \sqrt{7} - 1 \][/tex]
[tex]\[ \frac{3}{\sqrt{c}} = \sqrt{7} - 1 \][/tex]
6. Solve for [tex]\( c \)[/tex]:
Rearrange the equation to solve for [tex]\( c \)[/tex]:
[tex]\[ 3 = (\sqrt{7} - 1)\sqrt{c} \][/tex]
[tex]\[ \sqrt{c} = \frac{3}{\sqrt{7} - 1} \][/tex]
Squaring both sides to solve for [tex]\( c \)[/tex]:
[tex]\[ c = \left( \frac{3}{\sqrt{7} - 1} \right)^2 \][/tex]
7. Simplify the expression for [tex]\( c \)[/tex]:
Simplify the right side of the equation:
[tex]\[ c = \left( \frac{3}{\sqrt{7} - 1} \right)^2 = \frac{9}{(\sqrt{7} - 1)^2} \][/tex]
Therefore, the value of [tex]\( c \)[/tex] in the interval [tex]\( (1, 7) \)[/tex] that satisfies the Mean Value Theorem is:
[tex]\[ c = \frac{9}{(\sqrt{7} - 1)^2} \][/tex]
Thus, we have found that the [tex]\( c \)[/tex] such that [tex]\( f'(c) = \sqrt{7} - 1 \)[/tex] within the interval [tex]\( (1, 7) \)[/tex] is [tex]\( \frac{9}{(\sqrt{7} - 1)^2} \)[/tex].
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