# How do you find the average value of the function for #f(x)=sqrtx+1/sqrtx, 1<=x<=9#?

Use the formula

The FTC then leads us to say this equals

This simplifies to

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To find the average value of the function ( f(x) = \sqrt{x} + \frac{1}{\sqrt{x}} ) over the interval ( 1 \leq x \leq 9 ), you need to compute the definite integral of the function over that interval and then divide by the length of the interval.

The definite integral of ( f(x) ) over the interval ( [1, 9] ) is calculated as follows:

[ \int_{1}^{9} \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) , dx ]

After evaluating this integral, divide the result by the length of the interval, which is ( 9 - 1 = 8 ).

[ \text{Average value} = \frac{1}{8} \int_{1}^{9} \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) , dx ]

Once you evaluate the integral, you will have the average value of the function over the interval ( [1, 9] ).

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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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