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:

$y=\frac{1}{2-x}$

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

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

Substitute for y

$\begin{gathered} \Delta y=\frac{1}{2-(x+\Delta x)}-\frac{1}{2-x} \\ \Delta y=\frac{2-x-(2-(x+\Delta x)}{(2-(x+\Delta x)(2-x)} \\ \Delta y=\frac{2-x-(2-x-\Delta x)}{(2-(x+\Delta x)(2-x)} \\ \Delta y=\frac{2-x-2+x+\Delta x}{(2-(x+\Delta x)(2-x)} \\ \Delta y=\frac{\Delta x}{(2-(x+\Delta x)(2-x)} \end{gathered}$$\begin{gathered} \text{Divide through by }\Delta x \\ \frac{\Delta y}{\Delta x}=\frac{\Delta x}{(2-(x+\Delta x)(2-x)}\times\frac{1}{\Delta x} \\ \frac{\Delta y}{\Delta x}=\frac{1}{(2-(x+\Delta x)(2-x)} \end{gathered}$$\frac{dy}{dx}=\frac{1}{(2-x)(2-x)}$

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

$\frac{dy}{dx}=\frac{1}{(2-x)^2}$