Chapter 6.5, Problem 12E

To determine

To Find: The numbers b when the average value of $f\left(x\right)=2+6x-3x^2$ on the interval $\left[0,b\right]$.

Answer to Problem 12E

The value of b is $\underset{\_}{\frac{3\pm \sqrt{5}}{2}}$.

Explanation of Solution

Given information:

The function is $f\left(x\right)=2+6x-3x^2$ (1)

The requirement is average value of $f\left(x\right)$ is equal to 3.

Calculation:

The Expression to find the average value of function $\left(f_{\text{ave}}\right)$ on the interval.

$f_{\text{ave}}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ (2)

Substitute Equation (1) in (2).

$\begin{array}{c}\frac{1}{b-0}{\displaystyle \underset{0}{\overset{b}{\int }}\left(2+6x-3x^2\right)dx}=\frac{1}{b}{\left[2x+3x^2-x^3\right]}_0^b\\ =\frac{1}{b}\left[2b+3b^2-b^3-0\right]\\ =2+3b-b^2\end{array}$

Solve the equation.

$\begin{array}{c}2+3b-b^2=3\\ b^2-3b+1=0\end{array}$ (3)

Equation (3) is in the form of quadratic equation $a{x}^2+bx+c=0$.

To solve Equation (3) to find the value of b using the relation.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (4)

Substitute 1 for a, $-3$ for b, 1 for c and b for x in Equation (4).

$\begin{array}{c}b=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4\times 1\times 1}}{2\times 1}\\ =\frac{3\pm \sqrt{9-4}}{2}\\ =\frac{3\pm \sqrt{5}}{2}\end{array}$

Therefore, the value of b is $\underset{\_}{\frac{3\pm \sqrt{5}}{2}}$.