Question

My question is soon to follow...

Answer 1

I've done 7(i), but I'm having trouble with (ii)...

Answer 2

@DebbieG

Answer 3

I like it, cool problem. I'm on my way out but the gist of it:

find the inverse function for each portion of the piecewise function. And remember that the domain and ranges will "flip", so your inverse (which is a new piecewise function) will have domain that equates to the range of the original piece.

Answer 4

Okay, so y = 11 - x^2 inverted becomes y = + or - sqr(11-x), and y= 5-x inverted is itself.

Answer 5

hmmm so you apply the range of the original to the domain constrains of the new piecewise with the inverses

Answer 6

The domain of both originally is \[x \ge 0\] so then the range of both becomes \[x \ge0\]
And the domain of the first one is \[o \le x \le p\] and that of the second is \[x > p\]

Answer 7

well, the domain of \(\bf 11-x^2\) will be..... if set x = 0, then 11
if we set x = p, then \(\bf 11-p^2\) so the range will end up as \(\bf 11 \le x \le 11-p^2\)

Answer 8

hhhmmmm

Answer 9

\(11-q^2\) rather

Answer 10

shoot... can't be.... \(\bf 11-p^2 = q\)

notice the output, or range, goes down to coordinate "q"

Answer 11

\(\large { f^{-1}(x)\implies
\begin{cases}
\sqrt{11-x} \qquad& 11\le x \le q \\\quad \\
5-x\qquad & x < q
\end{cases}}\)

Answer 12

because, if you notice in f(x), those were the ranges for each equation
and thus
they become the domain in the inverse function

Answer 13

It's getting really late here, so I'm going to come back and have a look at this tomorrow, because its not making sense right now. :/ But thanks guys for the help. :)

Answer 14

\(\bf x = 0\\
11-(0)^2\implies 11\\ \quad \\
x = p\\
11-(p)^2 \implies q\qquad thus\\ \quad \\
\textit{the range for }11-x^2\implies\quad [11, q]\)

Answer 15

Right, I went through the question again, and this is what I got: \[f ^{-1}(x) = \sqrt{11 - x}\] with a domain of \[2 \le x \le 11\] and \[f ^{-1}(x) = 5 - x\] with a domain of \[x < 2\] From part (i) of the question, q = 2, so this looks correct to me, I think.

Answer 16

Thanks for the help, guys!