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avatar+3693 

If f(x) = xon the domain [-2,2] then f-1?

 

  • equals √x
  • has range [-2,2]
  • has domain [-4,4]
  • equals √x, has range [-2,2], and a domain of [-4,4]
  • none of the above
 Aug 23, 2016
 #1
avatar+128079 
0

f(x) = x^2    write f(x)  as y

 

y = x^2       get x by itself by taking the square root of both sides

 

√y =  x        "exchange"  x and y

 

√x = y         for y, write f-1(x)

 

√x  =  f-1(x)

 

In the original function, the domain is [-2, 2]  and the range is [0, 4]

 

In the inverse function,  the domain is  [0, 4]  and the range is [0, 2]

 

Here's a graph of both : https://www.desmos.com/calculator/rsveooxu3z

 

The first answer is the correct one  →  equals √x

 

BTW  -  good to see you.......how are you doing???

 

 

cool cool cool

 Aug 23, 2016
 #2
avatar+118587 
0

Hi Brittany, itr is REALLY good to see you again :))

 

Chris, I am not so sure about your answer.....

 

\(If\;\;y=x^2\qquad then\qquad x=\pm\sqrt y \)

 

If the domain of the original function was [0,2] then   \(x=+\sqrt y\)

 

And that would certainly have an inversese function.

 

BUT

given the original domain restriction of [-2,2]  the inverse becomes...

\(y=\pm\sqrt x \qquad \mbox{and this is not a function at all}\)

(It does not pass the vertical line test)

 

So I think, 

If f(x) = x2 on the domain [-2,2] then f-1 does not exist.

 

I could be wrong - I will certainly consider further discussion on the topic.  indecision

 Aug 24, 2016
 #3
avatar+3693 
+6

hi, guys!

I'm great. Its been such a long time since ive been on here. i miss it and the people on here too. Im fantastic. I am on my second yr of senior year in highschool because my dad didn't want to graduate me at 16 and send me to a university. Soooo....I'm taking another year of HS, dual enrolling at UMUC (University of Maryland University College), and then taking a gap year before I go to college. I also turn 17 in less than two months! woop!

 

yea its been crazy. 

BrittanyJ  Aug 24, 2016

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