I attached the problem, and im stumped, any help would be appreciated.......
D dwpelt New member Joined Oct 17, 2009 Messages 1 Oct 17, 2009 #1 I attached the problem, and im stumped, any help would be appreciated....... Attachments question 10.jpg 18.1 KB · Views: 29
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Oct 17, 2009 #2 Hello, dwpelt! I'll do the first one . . . For the function find the average rate of change of f\displaystyle ff for 1 to x.\displaystyle x.x. . . Formula: f(x)−f(1)x−1x≠1\displaystyle \text{Formula: }\:\frac{f(x) - f(1)}{x-1}\quad x \neq 1Formula: x−1f(x)−f(1)x=1 (1) f(x) = 3x+2\displaystyle (1)\;f(x) \:=\:\frac{3}{x+2}(1)f(x)=x+23 Click to expand... Do you understand the formula? It says: 1. Take f(x)2. Subtract f(1)3. Divide by x−1\displaystyle \text{It says: }\:\begin{array}{ccccc}\text{1. Take }f(x) \\\text{2. Subtract }f(1) \\ \text{3. Divide by }x-1 \end{array}It says: 1. Take f(x)2. Subtract f(1)3. Divide by x−1 1. The first step is easy; we have: f(x) = 3x+2\displaystyle \text{1. The first step is easy; we have: }\;f(x) \:=\:\frac{3}{x+2}1. The first step is easy; we have: f(x)=x+23 2. Subtract f(1)\displaystyle \text{2. Subtract }f(1)2. Subtract f(1) . . \(\displaystyle \text{First, find }f(1): }\;f(1) \:=\:\frac{3}{1+2} \:=\:\frac{3}{3} \:=\:1\) . . So we have: f(x)−f(1) = 3x+2−1\displaystyle \text{So we have: }\;f(x) - f(1) \;=\;\frac{3}{x+2} - 1So we have: f(x)−f(1)=x+23−1 . . Simplify: 3x+2−x+2x+2 = 3−(x+2)x+2 = 1−xx+2\displaystyle \text{Simplify: }\;\frac{3}{x+2} - \frac{x+2}{x+2} \;=\;\frac{3-(x+2)}{x+2} \;=\;\frac{1-x}{x+2}Simplify: x+23−x+2x+2=x+23−(x+2)=x+21−x 3. Divide by x−1\displaystyle \text{3. Divide by }x-13. Divide by x−1 . . 1x−1⋅1−xx+2 = −(x−1)(x−1)(x+2) = −1x+2\displaystyle \frac{1}{x-1}\cdot\frac{1-x}{x+2} \;=\;\frac{-(x-1)}{(x-1)(x+2)} \;=\;\boxed{\frac{-1}{x+2}}x−11⋅x+21−x=(x−1)(x+2)−(x−1)=x+2−1
Hello, dwpelt! I'll do the first one . . . For the function find the average rate of change of f\displaystyle ff for 1 to x.\displaystyle x.x. . . Formula: f(x)−f(1)x−1x≠1\displaystyle \text{Formula: }\:\frac{f(x) - f(1)}{x-1}\quad x \neq 1Formula: x−1f(x)−f(1)x=1 (1) f(x) = 3x+2\displaystyle (1)\;f(x) \:=\:\frac{3}{x+2}(1)f(x)=x+23 Click to expand... Do you understand the formula? It says: 1. Take f(x)2. Subtract f(1)3. Divide by x−1\displaystyle \text{It says: }\:\begin{array}{ccccc}\text{1. Take }f(x) \\\text{2. Subtract }f(1) \\ \text{3. Divide by }x-1 \end{array}It says: 1. Take f(x)2. Subtract f(1)3. Divide by x−1 1. The first step is easy; we have: f(x) = 3x+2\displaystyle \text{1. The first step is easy; we have: }\;f(x) \:=\:\frac{3}{x+2}1. The first step is easy; we have: f(x)=x+23 2. Subtract f(1)\displaystyle \text{2. Subtract }f(1)2. Subtract f(1) . . \(\displaystyle \text{First, find }f(1): }\;f(1) \:=\:\frac{3}{1+2} \:=\:\frac{3}{3} \:=\:1\) . . So we have: f(x)−f(1) = 3x+2−1\displaystyle \text{So we have: }\;f(x) - f(1) \;=\;\frac{3}{x+2} - 1So we have: f(x)−f(1)=x+23−1 . . Simplify: 3x+2−x+2x+2 = 3−(x+2)x+2 = 1−xx+2\displaystyle \text{Simplify: }\;\frac{3}{x+2} - \frac{x+2}{x+2} \;=\;\frac{3-(x+2)}{x+2} \;=\;\frac{1-x}{x+2}Simplify: x+23−x+2x+2=x+23−(x+2)=x+21−x 3. Divide by x−1\displaystyle \text{3. Divide by }x-13. Divide by x−1 . . 1x−1⋅1−xx+2 = −(x−1)(x−1)(x+2) = −1x+2\displaystyle \frac{1}{x-1}\cdot\frac{1-x}{x+2} \;=\;\frac{-(x-1)}{(x-1)(x+2)} \;=\;\boxed{\frac{-1}{x+2}}x−11⋅x+21−x=(x−1)(x+2)−(x−1)=x+2−1
