Question

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

Answer 1

Here, we must find the derivative of

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

Using the method of differentiation, we can solve it using the limit:

`f^(\prime)(x)=\lim _(\Delta x\rightarrow0)(f(x+\Delta x)-f(x))/(\Delta x)`

We can use the four-step rule, it's the same thing as calculating the limit, we just break it into more steps. First, we take the original expression and replace "x" with "x + Δx"

`f(x+\Delta x)=(1)/(2-(x+\Delta x))`

Now we subtract the original function from it

`\begin{gathered} f(x+\Delta x)-f(x)=(1)/(2-(x+\Delta x))-(1)/(2-x) \\ \\ f(x+\Delta x)-f(x)=\frac{(2-x)-(2-(x+\Delta x))_{}}{(2-(x+\Delta x))(2-x)} \\ \\ f(x+\Delta x)-f(x)=\frac{-\Delta x_{}}{(2-(x+\Delta x))(2-x)} \end{gathered}`

Now we divide both sides by Δx

`\begin{gathered} (f(x+\Delta x)-f(x))/(\Delta x)=(1)/(\Delta x)\cdot\frac{\Delta x_{}}{(2-(x+\Delta x))(2-x)} \\ \\ (f(x+\Delta x)-f(x))/(\Delta x)=\frac{1_{}}{(2-(x+\Delta x))(2-x)} \end{gathered}`

Now we can do the limit when Δx→0

`\begin{gathered} f^(\prime)(x)=\lim _(\Delta x\rightarrow0)(f(x+\Delta x)-f(x))/(\Delta x)=\lim _(\Delta x\rightarrow0)\frac{1_{}}{(2-(x+\Delta x))(2-x)} \\ \\ \lim _(\Delta x\rightarrow0)\frac{1_{}}{(2-(x+\Delta x))(2-x)}=(1)/((2-x)(2-x))=(1)/((2-x)^2) \end{gathered}`

Therefore the derivative is

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

But the exercise doesn't end here, we must find when the derivative is equal to 1/16, therefore:

`\begin{gathered} (1)/((2-x)^2)=(1)/(16) \\ \\ (2-x)^2=16 \\ \\ (2-x)^2=16 \end{gathered}`

Hence

`|2-x|=4\Rightarrow\begin{cases}2-x=4 \\ 2-x=-4\end{cases}\Rightarrow\begin{cases}x=-2 \\ x=6\end{cases}`

Therefore, the values of x that satisfies the problem are x = -2 and x = 6