Question

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

Answer 1

x = -2 and x = 6

Step-by-step explanation

The rate of change of the function is the derivative of the function.

In this case, we have the function,

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

Using the chain rule,

`f^(\prime)(u(x))=f^(\prime)(u)\cdot u^(\prime)(x)`

In this case, u = 2 - x and f(u) is u⁻¹,

`y^(\prime)=-1\cdot(2-x)^(-1-1)\cdot(-1)=(2-x)^(-2)`

The rate of change is,

`y^(\prime)=(1)/((2-x)^2)`

We have to find for which values of x, y' = 1/16. Thus, we have to solve the equation,

`(1)/(16)=(1)/((2-x)^2)`

Raise both sides to the exponent -1 - i.e. flip both sides of the equation,

`(2-x)^2=16`

Take the square root of both sides - remember that the square root has a negative and a positive result,

`\begin{gathered} \sqrt[]{(2-x)^2}=\pm\sqrt[]{16} \\ \\ 2-x=\pm4 \end{gathered}`

Subtract 2 from both sides,

`\begin{gathered} 2-2-x=-2\pm4 \\ \\ -x=-2\pm4 \end{gathered}`

And multiply both sides by -1,

`x=2\pm4`

Hence, the values of x for which the rate of change of y with respect to x is equal to 1/16 are x = -2 and x = 6