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The mean value theorem for the given function and interval. f(x)=√x-2 , [2,9]

User Mahmoh
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Final answer:

The mean value theorem states that for a function f(x) that is continuous on a closed interval and differentiable on an open interval, there exists at least one value c between the endpoints where the derivative is equal to the average rate of change of the function. In this case, the function is f(x) = √x - 2 and the interval is [2, 9]. By using the mean value theorem, we get c = 49/4.

Step-by-step explanation:

The mean value theorem states that for a function f(x) that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), there exists at least one value c between a and b such that the derivative of f(x) at c is equal to the average rate of change of f(x) over the interval [a, b].

In this case, the function is f(x) = √x - 2 and the interval is [2, 9].

To apply the mean value theorem, we need to check if the given function satisfies the conditions.

The function √x - 2 is continuous and differentiable for x > 2.

Therefore, we can use the mean value theorem to find the value of c.

First, let's find the derivative of f(x) = √x - 2.

The derivative is f'(x) = 1/(2√x).

Next, let's find the average rate of change of f(x) over the interval [2, 9].

The average rate of change is given by (f(9) - f(2))/(9 - 2).

Plugging in the values, we get (f(9) - f(2))/(9 - 2) = (√9 - 2 - (√2 - 2))/7 = (3 - 1)/7 = 2/7.

Now, we need to find the value of c between 2 and 9 such that f'(c) = 2/7.

Solving the equation 1/(2√c) = 2/7, we get c = 49/4.

Therefore, the mean value theorem guarantees that there exists a value c between 2 and 9 such that f'(c) = 2/7.

User I Am L
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