anonymous
  • anonymous
Inverse Functions and Their Graphs I Worksheet Part 1 please help!!! posting below.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I dont know if the stuff i did in red was right either...
anonymous
  • anonymous
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 \(\mathbb{R}\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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 \(y\geq 2\)
anonymous
  • anonymous
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.. ?
anonymous
  • anonymous
the domain and range are intervals, not single digits
anonymous
  • anonymous
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 \(y\geq 2\)
anonymous
  • anonymous
i will tell you what they want you to do to find \(f^{-1}\) if you tell me who wrote this "assessment"?
anonymous
  • anonymous
whats the domain then? 6?
anonymous
  • anonymous
no dear lets go slow the domain is not a number
anonymous
  • anonymous
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"
anonymous
  • anonymous
so its (−∞,∞)?
anonymous
  • anonymous
yes it is
anonymous
  • anonymous
that is true of any polynomial
anonymous
  • anonymous
oh okay
anonymous
  • anonymous
how do we find the inverse?
anonymous
  • anonymous
not sure how you write "all numbers greater than or equal to 2" for the range, but you could say \[y\geq 2\] or \[[2\infty)\]
anonymous
  • anonymous
sorry i meant \[[2,\infty)\]
anonymous
  • anonymous
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
anonymous
  • anonymous
Well i know teachers dont really write the assignments so whoever controls students education probably wrote
anonymous
  • anonymous
on line class?
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
what system?
anonymous
  • anonymous
huh?
anonymous
  • anonymous
FLVS? Keystone? k-12?
anonymous
  • anonymous
its just high school
anonymous
  • anonymous
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 \(y\)
anonymous
  • anonymous
\[x=2y^2+2\] subtract \(2\) get \[x-2=2y^2\] divide by b2 get \[\frac{x-2}{2}=y^2\] take the square root and get \[y=\pm\sqrt{\frac{x-2}{2}}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
tell them satellite73 says to get it together here
anonymous
  • anonymous
so thats the answer y= blah blah ?
anonymous
  • anonymous
yeah y = blah blah is right
anonymous
  • anonymous
domain of \(f^{-1} \)is \([2,\infty)\) same as the range of \(f\)
anonymous
  • anonymous
the range is \[Y \ge2\]
anonymous
  • anonymous
or is it 2,infinity
anonymous
  • anonymous
that is the domain of \(f^{-1}\) and the range of \(f\) same thing
anonymous
  • anonymous
it looks like this?
1 Attachment
anonymous
  • anonymous
last two are wrong
anonymous
  • anonymous
domain of \(f^{-1}\) is \([2,\infty)\) or \(y\geq 2\) whichever you like to write
anonymous
  • anonymous
okay .. and the range
anonymous
  • anonymous
ok i made a mistake there sorry domain is \([2,\infty)\) or \(x\geq 2\) x, not y
anonymous
  • anonymous
i dont get the range
anonymous
  • anonymous
i am not sure how you are supposed to answer that my guess is to say \((-\infty, \infty)\) but that is just a guess
anonymous
  • anonymous
how would you graph that? thats the 2nd part of it
anonymous
  • anonymous
which one, the inverse?
anonymous
  • anonymous
anonymous
  • anonymous
\[y=2x^2+2\] is a parabola that opens up with vertex at \((0,2)\)
anonymous
  • anonymous
looks like this http://www.wolframalpha.com/input/?i=y%3D2x^2%2B2
anonymous
  • anonymous
\[y=\pm\sqrt{\frac{x-2}{2}}\] is a parabola that opens to the right with vertex \((2,0)\)
anonymous
  • anonymous
so its 0,2 and 2,0?
anonymous
  • anonymous
and do we get the function out of the coordinates ?
anonymous
  • anonymous
that is the vertex take a look at the link i sent, shows you a picture of it
anonymous
  • anonymous
it doesnt say anything about the function?
anonymous
  • anonymous
|dw:1449714765597:dw|
anonymous
  • anonymous
im confused on the function part
anonymous
  • anonymous
the picture i just drew what the graph of each the one that opens up is \[y=2x^2+2\] the one that opens to the left is the inverse \[y=\pm\sqrt{\frac{x-2}{2}}\]
anonymous
  • anonymous
but.. i dont get how that plays in with the function
anonymous
  • anonymous
Hello @satellite73 ????
anonymous
  • anonymous
i still need help?????
anonymous
  • anonymous
i am not sure what you are asking i made a rough graph of the two functions above
anonymous
  • anonymous
its asking which is a function... do i say both??
anonymous
  • anonymous
oooh
anonymous
  • anonymous
the first one is \(f(x)=2x^2+2\) the second one \(f^{-1}\) is not
anonymous
  • anonymous
Okay thank you!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.