# 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?

Aug 1, 2017

Find the average value. Set the function equal to that value. Solve the equation in the interval. $c = \frac{1}{2} - \frac{\sqrt{21}}{6}$

#### Explanation:

Average value on $\left[- 1 , 1\right]$:

$\overline{y} = \frac{1}{1 - \left(- 1\right)} {\int}_{-} {1}^{1} \left({x}^{2} - x\right) \mathrm{dx}$

$= \frac{1}{2} {\left[{x}^{3} / 3 - {x}^{2} / 2\right]}_{-} {1}^{1}$

$= \frac{1}{3}$

Solve

${x}^{2} - x = \frac{1}{3}$

$3 {x}^{2} - 3 x - 1 = 0$

$x = \frac{3 \pm \sqrt{9 + 12}}{6}$

$= \frac{1}{2} \pm \frac{\sqrt{21}}{6}$

$5 < \sqrt{21} <$6, so $\frac{1}{2} < \frac{\sqrt{21}}{6} < 1$. Thus

$\frac{1}{2} + \frac{\sqrt{21}}{6}$ is outside the interval and

$\frac{1}{2} - \frac{\sqrt{21}}{6}$ is in the interval $\left[- 1 , 1\right]$.

$c = \frac{1}{2} - \frac{\sqrt{21}}{6}$