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202408201533 Example 13.8

Let \displaystyle{x=\frac{y-2}{y+2}}.

(a) Find \displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}} in terms of y.
(b) Make y the subject of the given function. Hence, find \displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}} in terms of x.
(c) Are the answers obtained in (a) and (b) different? Explain your assertion.

Extracted from S. W. Li et al. (2002). New Progress in Additional Mathematics Book 3

Roughwork.

\begin{aligned} x & = \frac{y-2}{y+2} \\ x & = \frac{y+2-4}{y+2} \\ \textrm{Eq. (1):}\quad x=f(y) & = 1-\frac{4}{y+2} \\ \frac{4}{y+2} & = 1-x \\ \frac{4}{1-x} & = y+2 \\ \textrm{Eq. (2):}\quad y=g(x) & = \frac{4}{1-x} - 2 \\ \end{aligned}

domain: what can go into a function
codomain: what may possibly come out of a function
range: what actually comes out of a function

Kevin Sookocheff on Range, Domain, and Codomain (2018)

Write, for function f,

\begin{aligned} \textrm{domain: }& \mathbb{R}\backslash\{ -2\} \\ \textrm{codomain: }& \mathbb{R} \\ \textrm{range: }& \mathbb{R} \\ \end{aligned}

and for function g,

\begin{aligned} \textrm{domain: }& \mathbb{R}\backslash\{ 1\} \\ \textrm{codomain: }& \mathbb{R} \\ \textrm{range: }& \mathbb{R} \\ \end{aligned}

We see f discontinuous at -2; and g, at 1; whereas non-differentiable.

(to be continued)