How can I determine this linear transformation?

Question

Let $a,b \in \mathbb R$, $a<b$ and let $f(x)=\sin^{-1}(x)$. Determine a linear transformation $g: \mathbb R\rightarrow\mathbb R$, $x\mapsto z_1x+z_0$ such that $g([a,b])=[-1,1]$. Specify the inverse function $g^{-1}$ and plot $f(g(x))$ where $x \in \mathbb [a,b]$.

I am not sure which transformation I am looking for here. My best guess is that this is referring to integration:

$$g[x]= \int dz [x]=z_1x+z_0$$

However, I am unsure how to interpret the statement that $g([a,b])=[-1,1]$. Would that mean that the definite integral from $a$ to $b$ would yield an area between $-1$ and $1$. Is integration even the correct transformation I am looking for here?

Answer 1

Take $g(x)=\dfrac{2x}{b-a}-\dfrac{a+b}{b-a}$. It is clear that:

• $g(a)=-1$;
• $g(b)=1$;
• $g(x)$ is of the form $z_1x+z_0$.

In fact, I computed $z_1$ and $z_0$ by solving the system$$\left\{\begin{array}{l}z_1a+z_0=-1\\z_1b+z_0=1.\end{array}\right.$$So, $g$ is the function that you're after.

Answer 2

Hint

If $g(z)=z_0+z_1x$, then $g$ is indeed affine but not linear (if $z_0\neq 0$). Anyway, if $g(z)=z_0+z_1x$ with $z_1\neq 0$, then $g$ is strictly monotone. i.e. $$g([a,b])=[-1,1]\iff \begin{cases}z_0+az_1=-1\text{ and }z_0+z_1b=1\\ \text{or}\\ z_0+az_1=1\text{ and }z_0+bz_1=-1\end{cases}.$$