Question

HELP ME PLEASE!!!!
graph each function and its inverse. after describe the domain and range of each.
f(x)=(x+3)^2

Answer 1

@jdoe0001

Answer 2

do you know how to graph \(\bf x^2\quad ?\)

Answer 3

well i have a graphing calculator so it does it for me. i dont know how to find the inverse and the domain and range. i know when you graph x^2 you get a u shape graph

Answer 4

Inverse of f(x)
\[
g(x)=f^{-1}(x) =\sqrt x-3
\]

Answer 5

so the inverse is \[\sqrt{x}-3\]

Answer 6

\(\begin{array}{llll}
f(x)=(x&+3)^2\\
&\uparrow \\
&\textit{horizontal shift of }x^2\textit{ graph to the left by 3 units}
\end{array}\)

Answer 7

yes

Answer 8

ok so how do i find the domain and range?

Answer 9

The domain of is all real numbers.
The domain of g is all real \( x \ge 0\)

Answer 10

to find the inverse "relationship" of an expression, you'd just swap the variables about \(\bf f(x)=\color{red}{y}=(x+3)^2\qquad inverse\implies x=(\color{red}{y}+3)^2\\ \quad \\
\sqrt{x}=y+3\implies \sqrt{x}-3=y\)

and solve for "y"

Answer 11

ok i dont understand the domain and range still

Answer 12

The domain is all x so that the function evaluated at x makes sense

Answer 13

The range of a function h is all the values of h(x)

Answer 14

the domain of a function is, all values that "x" can take, that is, the INPUT
meaning, any values for "x" that produce a valid "y", not an undefined or extraneous values

so for \(\bf f(x)=y=(x+3)^2\) "x" can pretty much take anything, thus \(\bf x \in \mathbb{R}\)

Answer 15

@jdoe0001 can help you finish this problem

Answer 16

for \(\bf \sqrt{x}-3=y\) the radical CANNOT be negative, because when it's negative you'd end up with an imaginary value, thus not a valid one

so for \(\bf \sqrt{x}-3=y\) "x" cannot be negative, that is, has to always be either positive or 0 at the very least

Answer 17

ok that will work but you lost me on the last part im so confused on the values of x and them being real numbers and all that. what is xr thing

Answer 18

so the domain is 0 and up for the inverse?

Answer 19

\(\begin{array}{llll}
\mathbb{R} \implies \textit{all real numbers set}\\ \quad \\
\bf x\in \mathbb{R}\implies \textit{"x" belongs to all real numbers set}
\end{array}\)

Answer 20

0 and up, yes, I assume you've covered "imaginary numbers", so when you have a negative radicand, that's what you've got

Answer 21

yes. so after that how do you find the range of the two functions

Answer 22

..... well... notice your graph in the calculator for \(\bf f(x)=(x+3)^2\) what's its vertex?

Answer 23

the vertex is -3??? i think

Answer 24

-3? x= -3? y = -3?

Answer 25

what's the (x,y) pair for the vertex?

Answer 26

oh i forgot to put it in that form it is (-3,0)

Answer 27

does the graph go below (-3, 0)? no, since that's the vertex, meaning that "y" values don't go pass 0, they do go ABOVE 0, not BELOW 0

so that's the range
range= values for "y"

Answer 28

you can also try graphing maybe \(\bf y=\sqrt{x}-3\) and you'll also see the range as well, by just looking at the graph

Answer 29

so both x and y are values greater than or equal to 0 correct

Answer 30

is that the domain and range for both functions?

Answer 31

well.... for \(\bf f(x)=y=(x+3)^2\) the values for "y" don't go pass the vertex of (-3, 0)
so the range goes ABOVE 0, not BELOW 0, that'd be the range for that one
recall that for "x" or the domain was \(\bf x \in \mathbb{R}\)

Answer 32

"so the range goes ABOVE 0, not BELOW 0" <---- so, what does that make the range?