How do you find the average value of the function for #f(x)=(3x)/sqrt(1x^2), 1/2<=x<=1/2#?
The average value is
So our equation will be
graph{y = (3x)/sqrt(1  x^2) [10, 10, 5, 5]}
Hopefully this helps!
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To find the average value of the function ( f(x) = \frac{3x}{\sqrt{1x^2}} ) over the interval (\frac{1}{2} \leq x \leq \frac{1}{2}), you need to evaluate the definite integral of (f(x)) over that interval and then divide it by the length of the interval.

First, compute the definite integral of (f(x)) over the given interval: [ \int_{1/2}^{1/2} \frac{3x}{\sqrt{1x^2}} , dx ]

Next, evaluate this integral.

Then, find the length of the interval, which is ( \frac{1}{2}  (\frac{1}{2}) = 1 ).

Finally, divide the result from step 2 by the length of the interval (step 3) to find the average value of the function over the interval.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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