Question

find the numbers b such that the average value of $f(x)=2 + 10x-6x^{2}$ on the interval $0,b$ is equal to 3.

b = (smaller value)
b = (larger value)

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $\bar{y}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 0$, $f(x)=2 + 10x-6x^{2}$, and $\bar{y}=3$. So we have the equation $\frac{1}{b-0}\int_{0}^{b}(2 + 10x-6x^{2})dx=3$.

Step2: Calculate the integral

First, integrate $2 + 10x-6x^{2}$ with respect to $x$. $\int(2 + 10x-6x^{2})dx=2x + 5x^{2}-2x^{3}+C$. Then $\int_{0}^{b}(2 + 10x-6x^{2})dx=(2b + 5b^{2}-2b^{3})-(2\times0 + 5\times0^{2}-2\times0^{3})=2b + 5b^{2}-2b^{3}$.

Step3: Substitute into the average - value equation

The equation $\frac{1}{b}\int_{0}^{b}(2 + 10x-6x^{2})dx = 3$ becomes $\frac{2b + 5b^{2}-2b^{3}}{b}=3$ for $b
eq0$. Simplify the left - hand side: $2 + 5b-2b^{2}=3$.

Step4: Rearrange to a quadratic equation

Rearrange $2 + 5b-2b^{2}=3$ to $2b^{2}-5b + 1=0$.

Step5: Use the quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For $2b^{2}-5b + 1=0$, where $a = 2$, $b=-5$, and $c = 1$, we have $b=\frac{5\pm\sqrt{(-5)^{2}-4\times2\times1}}{2\times2}=\frac{5\pm\sqrt{25 - 8}}{4}=\frac{5\pm\sqrt{17}}{4}$.

Step6: Identify smaller and larger values

The two solutions for $b$ are $b_1=\frac{5-\sqrt{17}}{4}\approx\frac{5 - 4.123}{4}=\frac{0.877}{4}=0.219$ and $b_2=\frac{5+\sqrt{17}}{4}\approx\frac{5 + 4.123}{4}=\frac{9.123}{4}=2.281$.

Answer:

Smaller value: $\frac{5-\sqrt{17}}{4}$
Larger value: $\frac{5+\sqrt{17}}{4}$