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Problem 122

# Obtain an approximate value for $$\sqrt{105}$$ to within $$.01$$ by using the Mean Value Theorem.

Expert verified
Using the Mean Value Theorem on the function $$f(x) = x^2$$, we find an estimate of the desired value within the interval [10, 11] to be 10.5. This estimate is not accurate to within 0.01, but it can be used as a starting point for more refined approximations.
See the step by step solution

## Step 1: Define the function representing the square root.

Let's use the function $$f(x)=x^2$$, where $$x^2=105$$. We want to find the value of x such that $$x^2$$ is 105.

## Step 2: Determine the interval containing the desired value.

We know that 10^2 = 100 and 11^2 = 121. Since 105 is between 100 and 121, the desired value must be between 10 and 11. So, the interval we will work on is [10, 11].

## Step 3: Apply the Mean Value Theorem.

Applying the Mean Value Theorem to the function $$f(x)=x^2$$ on the interval [10, 11], we have: $$f'(c) = \frac{f(11) - f(10)}{11 - 10}$$ for some c in (10, 11), where f'(c) denotes the derivative of the function with respect to x. Now, let's find the derivative of $$f(x) = x^2$$: $$f'(x) = 2x$$ We can now plug this into our Mean Value Theorem equation: $$2c = \frac{(11)^2 - (10)^2}{11 - 10}$$ Solve for c: $$2c = \frac{121 - 100}{1}$$ $$2c = 21$$ $$c = 10.5$$ Using the Mean Value Theorem, we've found that 10.5 is an estimate of the desired value, which should be to within 0.01.

## Step 4: Check the accuracy of the approximation.

We can verify that our estimate is accurate within 0.01 by computing the square of our estimate, which should be close enough to 105. $$(10.5)^2 = 110.25$$ Our estimate is within 5.25 of 105, which is larger than the desired level of accuracy (0.01). However, we can use the estimate, 10.5, as the starting point for a more refined approximation, such as using the bisection method or Newton's method, to obtain a more accurate value of the square root of 105.

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