anonymous
  • anonymous
Imagine, there's a cubic graph which has already been plotted, and imagine they want you to graphically plot it's inverse. In which line do you have to reflect it??
Mathematics
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anonymous
  • anonymous
Imagine, there's a cubic graph which has already been plotted, and imagine they want you to graphically plot it's inverse. In which line do you have to reflect it??
Mathematics
chestercat
  • chestercat
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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.

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anonymous
  • anonymous
Do u mean this graph?
anonymous
  • anonymous
|dw:1337492052470:dw|
anonymous
  • anonymous
I mean, literally ANY graph :P

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anonymous
  • anonymous
So, is my graph correct?
anonymous
  • anonymous
yeahh, i guess :P wait is this the cubic graph, or the line in which it hsa to be reflected in, or the reflected cubic graph?? i'm confused :P
apoorvk
  • apoorvk
The graph of inverse of any function is just the same graph that is its mirror image about the 'y=x' line - basically flipped about the 'diagonal'.
apoorvk
  • apoorvk
(ofcourse provided that the function if bijective and the inverse exists)
apoorvk
  • apoorvk
Yeah, and so in you y=x^3 graph, the curve of the inverse is simply flipped about 'y=x' something like this: (I think I need to draw a new curve): |dw:1337492674235:dw|
apoorvk
  • apoorvk
|dw:1337492883274:dw| That's^^ the inverse of y=x^3, or basically the graph of y=x^(1/3) (in black)
anonymous
  • anonymous
Yeah. @apoorvk is right. Yesterday, our maths sir taught this. Thanks Tanvi for helping me to revise my concepts. :P

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