# 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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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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