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

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

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

$f^{\prime}(x)=\lim _{\Delta x\rightarrow0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

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"

$f(x+\Delta x)=\frac{1}{2-(x+\Delta x)}$

Now we subtract the original function from it

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

Now we divide both sides by Δx

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

Now we can do the limit when Δx→0

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

Therefore the derivative is

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

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

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

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