1. ## Transformations

Given the parent functions;
$\displaystyle f(x)=c$
$\displaystyle f(x)=x$
$\displaystyle f(x)=x^2$
$\displaystyle f(x)=x^3$
$\displaystyle f(x)=x^n$
$\displaystyle f(x)=|x|$
$\displaystyle f(x)=b^x$
$\displaystyle f(x)=\sqrt{x}$
$\displaystyle f(x)=1/x$
$\displaystyle f(x)=[x]$

I'm asked to rewrite these functions in terms of thier parent function. I've shown what I've done, and marked ones that I'm pretty sure I messed up on. If anyone could buzz over these and still be willing to hand out a couple pointers, that would be awesome.

$\displaystyle g(x)=3|x-2| => g(x)= 3f(x-2), ||||where f(x)=x$
$\displaystyle g(x)=-x^2+4 => g(x)= -f(x)+4, ||||where f(x)=x^2$
$\displaystyle *g(x)=(x-4)^3 => g(x)= f((x-4)^3), ||||where f(x)=x^3$
$\displaystyle *g(x)=3^{-x}-2 => g(x)= 3f(-x)-2, ||||where f(x)=b^x$
$\displaystyle *g(x)=3/x+2 => g(x)=3/f(x+2), ||||where f(x)=1/x$
$\displaystyle g(x)=-[2x] => g(x)=-f(2x), ||||where f(x)=[x]$
$\displaystyle g(x)=-4\sqrt{x-2} => g(x)=-4f(x-2), ||||where f(x)=\sqrt{x}$

*I'm not sure If I've done these ones right.

2. Hello, EyesForEars!

Given the parent functions:

. . $\displaystyle \begin{array}{ccccc}f(x)\,=\,c &f(x),=\,x & f(x)\,=\,x^2 & f(x)\,=\,x^3 & f(x)\,=\,x^n \\ \\[-3mm] f(x)\,=\,|x| & f(x)\,=\,b^x & f(x)\,=\,\sqrt{x} & f(x)\,=\,\frac{1}{x} & f(x)\,=\,[x] \end{array}$

I'm asked to rewrite these functions in terms of thier parent function.
I've shown what I've done, and marked ones that I'm pretty sure I messed up on.

$\displaystyle g(x)\,=\,3|x-2|\quad\Rightarrow\quad g(x)\,=\, 3f(x-2)\;\text{ where }f(x)\,=\,x$

$\displaystyle g(x)\,=\,-x^2+4\quad\Rightarrow\quad g(x)\,=\, -f(x)+4\;\text{ where }f(x)\,=\,x^2$

$\displaystyle g(x)\,=\,(x-4)^3 \quad\Rightarrow\quad g(x)\,=\, f((x-4)^3)\;\text{ where }f(x)\,=\,x^3$ .. . . no

$\displaystyle g(x)\,=\,3^{-x}-2 \quad\Rightarrow\quad g(x)\,=\, 3f(-x)-2\;\text{ where }f(x)\,=\,b^x$ . . . . Right!

$\displaystyle g(x)\,=\,\frac{3}{x+2} \quad\Rightarrow\quad g(x)\,=\,\frac{3}{f(x+2)}\;\text{ where }f(x)\,=\,1/x$ . . . . no

$\displaystyle g(x)\,=\,-[2x] \quad\Rightarrow\quad g(x)\,=\,-f(2x)\;\text{ where }f(x)\,=\,[x]$

$\displaystyle g(x)\,=\,-4\sqrt{x-2} \quad\Rightarrow\quad g(x)\,=\,-4f(x-2)\;\text{ where }f(x)\,=\,\sqrt{x}$

If $\displaystyle f(x) \:=\:x^3$, then $\displaystyle f(x-4) \:=\:(x-4)^3$

. . Therefore: .$\displaystyle g(x) \:=\:(x-4)^3 \:=\:f(x-4)$

If $\displaystyle f(x) \:=\:\frac{1}{x}$, then $\displaystyle f(x+2) \:=\:\frac{1}{x+2}$
. . Therefore: .$\displaystyle g(x) \:=\:\frac{3}{x+2} \:=\:3f(x+2)$

3. Awesome.