Question

for the function y=1/2-x at what values of x will the rate of change of y with respect to x equal 1/16

Answer 1

Given:

`y=(1)/(2-x)`

To Determine: Using the increament method the rate of change of y with respect to x

`\begin{gathered} y+\Delta y=(1)/(2-(x+\Delta x)) \\ \Delta y=(1)/(2-(x+\Delta x))-y \end{gathered}`

Substitute for y

`\begin{gathered} \Delta y=(1)/(2-(x+\Delta x))-(1)/(2-x) \\ \Delta y=(2-x-(2-(x+\Delta x))/((2-(x+\Delta x)(2-x)) \\ \Delta y=(2-x-(2-x-\Delta x))/((2-(x+\Delta x)(2-x)) \\ \Delta y=(2-x-2+x+\Delta x)/((2-(x+\Delta x)(2-x)) \\ \Delta y=(\Delta x)/((2-(x+\Delta x)(2-x)) \end{gathered}`
`\begin{gathered} \text{Divide through by }\Delta x \\ (\Delta y)/(\Delta x)=(\Delta x)/((2-(x+\Delta x)(2-x))*(1)/(\Delta x) \\ (\Delta y)/(\Delta x)=(1)/((2-(x+\Delta x)(2-x)) \end{gathered}`
`(dy)/(dx)=(1)/((2-x)(2-x))`

Hence, the rate of change of y with respect to x is

`(dy)/(dx)=(1)/((2-x)^2)`