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 \(\displaystyle f\) for 1 to \(\displaystyle x.\) . . \(\displaystyle \text{Formula: }\:\frac{f(x) - f(1)}{x-1}\quad x \neq 1\) \(\displaystyle (1)\;f(x) \:=\:\frac{3}{x+2}\) Click to expand... Do you understand the formula? \(\displaystyle \text{It says: }\:\begin{array}{ccccc}\text{1. Take }f(x) \\\text{2. Subtract }f(1) \\ \text{3. Divide by }x-1 \end{array}\) \(\displaystyle \text{1. The first step is easy; we have: }\;f(x) \:=\:\frac{3}{x+2}\) \(\displaystyle \text{2. Subtract }f(1)\) . . \(\displaystyle \text{First, find }f(1): }\;f(1) \:=\:\frac{3}{1+2} \:=\:\frac{3}{3} \:=\:1\) . . \(\displaystyle \text{So we have: }\;f(x) - f(1) \;=\;\frac{3}{x+2} - 1\) . . \(\displaystyle \text{Simplify: }\;\frac{3}{x+2} - \frac{x+2}{x+2} \;=\;\frac{3-(x+2)}{x+2} \;=\;\frac{1-x}{x+2}\) \(\displaystyle \text{3. Divide by }x-1\) . . \(\displaystyle \frac{1}{x-1}\cdot\frac{1-x}{x+2} \;=\;\frac{-(x-1)}{(x-1)(x+2)} \;=\;\boxed{\frac{-1}{x+2}}\)
Hello, dwpelt! I'll do the first one . . . For the function find the average rate of change of \(\displaystyle f\) for 1 to \(\displaystyle x.\) . . \(\displaystyle \text{Formula: }\:\frac{f(x) - f(1)}{x-1}\quad x \neq 1\) \(\displaystyle (1)\;f(x) \:=\:\frac{3}{x+2}\) Click to expand... Do you understand the formula? \(\displaystyle \text{It says: }\:\begin{array}{ccccc}\text{1. Take }f(x) \\\text{2. Subtract }f(1) \\ \text{3. Divide by }x-1 \end{array}\) \(\displaystyle \text{1. The first step is easy; we have: }\;f(x) \:=\:\frac{3}{x+2}\) \(\displaystyle \text{2. Subtract }f(1)\) . . \(\displaystyle \text{First, find }f(1): }\;f(1) \:=\:\frac{3}{1+2} \:=\:\frac{3}{3} \:=\:1\) . . \(\displaystyle \text{So we have: }\;f(x) - f(1) \;=\;\frac{3}{x+2} - 1\) . . \(\displaystyle \text{Simplify: }\;\frac{3}{x+2} - \frac{x+2}{x+2} \;=\;\frac{3-(x+2)}{x+2} \;=\;\frac{1-x}{x+2}\) \(\displaystyle \text{3. Divide by }x-1\) . . \(\displaystyle \frac{1}{x-1}\cdot\frac{1-x}{x+2} \;=\;\frac{-(x-1)}{(x-1)(x+2)} \;=\;\boxed{\frac{-1}{x+2}}\)
