# How do you find the point c in the interval #-1<=x<=1# such that f(c) is equation to the average value of #f(x)=x^2-x#?

Find the average value. Set the function equal to that value. Solve the equation in the interval.

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To find the point ( c ) in the interval (-1 \leq x \leq 1) such that ( f(c) ) is equal to the average value of ( f(x) = x^2 - x ), you would first calculate the average value of ( f(x) ) over the interval ([-1, 1]) using the formula:

[ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) , dx ]

where ( a ) and ( b ) are the lower and upper bounds of the interval, respectively. Then, you would set ( f(c) ) equal to this average value and solve for ( c ).

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