# How do you find the point c in the interval 0<=x<=3 such that f(c) is equation to the average value of f(x)=3x^2?

Aug 6, 2017

Find the average value. Set $f \left(x\right)$ equal to the average value. Solve that equation in the interval $\left[0 , 3\right]$.

#### Explanation:

${f}_{\text{ave}} = \frac{1}{3 - 0} {\int}_{0}^{3} 3 {x}^{2} \mathrm{dx}$

$= \frac{1}{3} {\left[{x}^{3}\right]}_{0}^{3}$

$\frac{1}{3} \left[\left({3}^{3}\right) - \left({0}^{3}\right)\right] = \frac{1}{3} \left(27\right) = 9$

Now solve

$3 {x}^{2} = 9$ $\text{ }$ in $\left[0 , 3\right]$

${x}^{2} = 3$

$x = \pm \sqrt{3}$ $\text{ }$ Note that the algebra gives us two solutions, but one of them is not in the interval.

$c = \sqrt{3}$