Question

Inverse Functions and Their Graphs I Worksheet Part 1
please help!!!
posting below.

Answer 1

I dont know if the stuff i did in red was right either...

Answer 2

lord help us
a) \[f(x)=2x^2+2\] is not a one to one function, so it does not have an inverse
b) you do not "find" the domain, you are told the domain
but if for some reason you are not told, you can assume the domain of any polynomial is all real numbers \(\mathbb{R}\)

Answer 3

the range we can find
since \(x^2\geq 0\) always, (since it is a square) you know \(2x^2+2\geq 2\) the range is \(y\geq 2\)

Answer 4

http://www.wolframalpha.com/input/?i=f%28x%29%3D2x2%2B2
this is the correct range and domain? And what i wrote for range was wrong.. ?

Answer 5

the domain and range are intervals, not single digits

Answer 6

wolfram is not telling you what the range is
the fact that the curve is always greater than or equal to 2 means the range is \(y\geq 2\)

Answer 7

i will tell you what they want you to do to find \(f^{-1}\) if you tell me who wrote this "assessment"?

Answer 8

whats the domain then? 6?

Answer 9

no dear lets go slow
the domain is not a number

Answer 10

the domain is an interval, a set of numbers
'in this case, since you have a polynomial, the domain is all numbers
all real numbers
\((-\infty, \infty)\)
\(\mathbb{R}\) however you write "all real numbers"

Answer 11

so its (−∞,∞)?

Answer 12

yes it is

Answer 13

that is true of any polynomial

Answer 14

oh okay

Answer 15

how do we find the inverse?

Answer 16

not sure how you write "all numbers greater than or equal to 2" for the range, but you could say \[y\geq 2\] or \[[2\infty)\]

Answer 17

sorry i meant \[[2,\infty)\]

Answer 18

i will tell you what they want for the inverse after you tell me exactly where this question came from
i am not interested in whether it is a test or not, i want to know who wrote it

Answer 19

Well i know teachers dont really write the assignments so whoever controls students education probably wrote

Answer 20

on line class?

Answer 21

Yeah

Answer 22

what system?

Answer 23

huh?

Answer 24

FLVS?
Keystone?
k-12?

Answer 25

its just high school

Answer 26

ok i am done torturing you
lets find the inverse
put \[y=2x^2+2\] the switch x and y because that is what the inverse does, write \[x=2y^2+2\] and solve for \(y\)

Answer 27

\[x=2y^2+2\] subtract \(2\) get \[x-2=2y^2\] divide by b2 get \[\frac{x-2}{2}=y^2\] take the square root and get \[y=\pm\sqrt{\frac{x-2}{2}}\]

Answer 28

that is the inverse
but it is NOT a function because of the \(\pm\) which annoys me because it means that the person who wrote this does not know what they are doing and should stop writing math questions

Answer 29

tell them satellite73 says to get it together here

Answer 30

so thats the answer y= blah blah ?

Answer 31

yeah y = blah blah is right

Answer 32

domain of \(f^{-1} \)is \([2,\infty)\) same as the range of \(f\)

Answer 33

the range is \[Y \ge2\]

Answer 34

or is it 2,infinity

Answer 35

that is the domain of \(f^{-1}\) and the range of \(f\)
same thing

Answer 36

it looks like this?

Answer 37

last two are wrong

Answer 38

domain of \(f^{-1}\) is \([2,\infty)\) or \(y\geq 2\) whichever you like to write

Answer 39

okay .. and the range

Answer 40

ok i made a mistake there sorry
domain is \([2,\infty)\) or \(x\geq 2\)
x, not y

Answer 41

i dont get the range

Answer 42

i am not sure how you are supposed to answer that
my guess is to say \((-\infty, \infty)\) but that is just a guess

Answer 43

how would you graph that? thats the 2nd part of it

Answer 44

which one, the inverse?

Answer 45

\[y=2x^2+2\] is a parabola that opens up with vertex at \((0,2)\)

Answer 46

looks like this
http://www.wolframalpha.com/input/?i=y%3D2x^2%2B2

Answer 47

\[y=\pm\sqrt{\frac{x-2}{2}}\] is a parabola that opens to the right with vertex \((2,0)\)

Answer 48

so its 0,2 and 2,0?

Answer 49

and do we get the function out of the coordinates ?

Answer 50

that is the vertex
take a look at the link i sent, shows you a picture of it

Answer 51

it doesnt say anything about the function?