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

ANSWER

x = -2 and x = 6

EXPLANATION

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

In this case, we have the function,

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

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,

$y^{\prime}=\frac{1}{(2-x)^2}$

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

$\frac{1}{16}=\frac{1}{(2-x)^2}$

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