22. find the numbers b such that the average value of f(x)=2 + 6x - 3x² on the interval 0, b is equal to 3.

Explanation:

Step1: Recall the average value formula

The average value of a function (y = f(x)) on the interval ([a,b]) is given by (f_{avg}=\frac{1}{b - a}\int_{a}^{b}f(x)dx). Here (a = 0), (f(x)=2 + 6x-3x^{2}), and (f_{avg}=3). So we have (3=\frac{1}{b-0}\int_{0}^{b}(2 + 6x-3x^{2})dx).

Step2: Compute the integral

First, find (\int(2 + 6x-3x^{2})dx). Using the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we get (\int(2 + 6x-3x^{2})dx=2x+3x^{2}-x^{3}+C). Then (\int_{0}^{b}(2 + 6x-3x^{2})dx=\left[2x+3x^{2}-x^{3}\right]_{0}^{b}=2b + 3b^{2}-b^{3}).

Step3: Substitute the integral result into the average - value formula

Substitute into (3=\frac{1}{b}\int_{0}^{b}(2 + 6x-3x^{2})dx), we have (3=\frac{2b + 3b^{2}-b^{3}}{b}) (for (b\neq0)). Since (b\neq0), we can cancel out (b) in the right - hand side (when (b = 0), the interval ([0,b]) is not well - defined). So (3 = 2+3b - b^{2}).

Step4: Rearrange the equation

Rearrange (3 = 2+3b - b^{2}) to the standard quadratic form (b^{2}-3b + 1=0).

Step5: Solve the quadratic equation

Using the quadratic formula (x=\frac{-B\pm\sqrt{B^{2}-4AC}}{2A}) for the quadratic equation (Ax^{2}+Bx + C = 0). Here (A = 1), (B=-3), (C = 1). Then (b=\frac{3\pm\sqrt{(-3)^{2}-4\times1\times1}}{2\times1}=\frac{3\pm\sqrt{9 - 4}}{2}=\frac{3\pm\sqrt{5}}{2}).

Answer:

(b=\frac{3+\sqrt{5}}{2}) or (b=\frac{3-\sqrt{5}}{2})