Consider the function f(x)=8(x^(1/2))+7 on the interval [1,5].
By the Mean Value Theorem, we know there exists a c in the open interval (1,5) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.
f(5) = 7 + 8 sqrt 5
f(1) = 7 + 8 sqrt 1
f(5) - f(1) = 8(sqrt 5 - 1) = 9.89
9.89/4 = mean slope = 2.47
so where does f'(x) = 2.47 ????
f'(x) = 4 /x^(1/2)
so
2.47 = 4/x^.5
x^.5 = 4/2.47 = 1.62
x = 2.62
check that please.
To apply the Mean Value Theorem to the function \( f(x) = 8\sqrt{x} + 7 \) on the interval \([1, 5]\), we need to check two conditions:
1. The function must be continuous on the closed interval \([1, 5]\). The function \( f(x) = 8\sqrt{x} + 7 \) is continuous on this interval since it is a polynomial.
2. The function must be differentiable on the open interval \((1, 5)\). The derivative of \( f(x) \) is \( f'(x) = \frac{4}{\sqrt{x}} \), which is defined and continuous on the open interval \((1, 5)\). So, the function is differentiable in this interval.
Now, to find the value of \( c \) that satisfies the conditions of the Mean Value Theorem, we need to find the mean slope of the function on the interval \([1, 5]\). The mean slope is given by the formula:
\( \frac{f(5) - f(1)}{5 - 1} \)
First, let's find \( f(5) \):
\( f(5) = 8\sqrt{5} + 7 \)
Next, let's find \( f(1) \):
\( f(1) = 8\sqrt{1} + 7 \)
Now, substitute these values into the formula for the mean slope:
\( \frac{f(5) - f(1)}{5 - 1} = \frac{(8\sqrt{5} + 7) - (8\sqrt{1} + 7)}{4} \)
Simplifying the expression gives us:
\( \frac{8\sqrt{5} - 8\sqrt{1}}{4} = \frac{8(\sqrt{5} - \sqrt{1})}{4} = 2(\sqrt{5} - 1) \)
This expression represents the mean slope on the interval \([1, 5]\).
According to the Mean Value Theorem, there exists a \( c \) in the open interval \((1, 5)\) such that \( f'(c) \) is equal to the mean slope. Therefore, to find the value of \( c \), we need to solve the following equation:
\( f'(c) = 2(\sqrt{5} - 1) \)
The derivative of \( f(x) = 8\sqrt{x} + 7 \) is \( f'(x) = \frac{4}{\sqrt{x}} \). So, we have:
\( \frac{4}{\sqrt{c}} = 2(\sqrt{5} - 1) \)
To solve for \( c \), we can multiply both sides of the equation by \( \sqrt{c} \) and then divide both sides by \( 2(\sqrt{5} - 1) \):
\( \sqrt{c} = \frac{4}{2(\sqrt{5} - 1)} \)
Simplifying the expression gives us:
\( \sqrt{c} = \frac{4}{2\sqrt{5} - 2} \)
To remove the square root, we can square both sides of the equation:
\( c = \left(\frac{4}{2\sqrt{5} - 2}\right)^2 \)
Evaluating the expression gives us:
\( c \approx 3.577 \)
Therefore, the value of \( c \) that satisfies the conditions of the Mean Value Theorem is approximately 3.577.