# How do you use the Mean Value Theorem to solve #f(x)=x - sqrtx# on the closed interval [0,4]?

I assume in your question that you desire to find a value

A comment about what we are really doing here: we are not really "using" the Mean Value Theorem to solve anything. Rather, we are solving an algebra equation related to the instantaneous rate of change

Here's a picture of the situation in this problem. Note that the two dashed lines have equal slopes:

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To solve ( f(x) = x - \sqrt{x} ) on the closed interval ([0,4]) using the Mean Value Theorem:

- Verify that the function ( f(x) ) is continuous on the closed interval ([0,4]) and differentiable on the open interval ((0,4)).
- Calculate the derivative of ( f(x) ), denoted as ( f'(x) ).
- Find the average rate of change of ( f(x) ) over the interval ([0,4]) by evaluating ( \frac{f(4) - f(0)}{4 - 0} ).
- Apply the Mean Value Theorem, which states that there exists at least one ( c ) in the open interval ((0,4)) such that ( f'(c) = \frac{f(4) - f(0)}{4 - 0} ).
- Solve for ( c ) by setting ( f'(c) ) equal to the average rate of change found in step 3.
- Once you find the value of ( c ), evaluate ( f(c) ) to find the point on the curve where the tangent line is parallel to the secant line.

By following these steps, you can use the Mean Value Theorem to solve ( f(x) = x - \sqrt{x} ) on the closed interval ([0,4]).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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