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vera_ewing

  • one year ago

Pre-calculus question

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  1. vera_ewing
    • one year ago
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  2. ribhu
    • one year ago
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    its B...

  3. ribhu
    • one year ago
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    replace f(x) by y and x by f inverse x.... and then solve.. you will get option B

  4. ganeshie8
    • one year ago
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    A function must have exactly one "y value" for a given "x value". Look at the options, all of them are of form \(y=\pm \text{somthing}\), which means, the \(y\) values are not unique. What can you infer fromm this ?

  5. vera_ewing
    • one year ago
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    It is not a function, because there are more than one y value?

  6. ganeshie8
    • one year ago
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    Exactly!

  7. ribhu
    • one year ago
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    see when a function has a non zero derivative at a point then it is invertible in the neighbourhood

  8. ganeshie8
    • one year ago
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    Answer is either A or C

  9. ribhu
    • one year ago
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    and if it would not be a function then how can you calculate inverse of it..

  10. ribhu
    • one year ago
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    i think it is a function. and answer should be B.

  11. ganeshie8
    • one year ago
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    good question, looks they don't care about these fine details at school level

  12. vera_ewing
    • one year ago
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    Well I'm pretty sure the answer is C...

  13. ribhu
    • one year ago
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    so avoiding any complexity to the best of the data provided the answer should be B for sure

  14. ganeshie8
    • one year ago
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    I bet my 10 cents on A or C

  15. vera_ewing
    • one year ago
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    ribhu, I think it is C.

  16. vera_ewing
    • one year ago
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    Ok, thanks for your help!! @ganeshie8 and @ribhu

  17. ganeshie8
    • one year ago
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    I think, we may say the inverse of given function is a multivalued function. But, we cannot say the inverse is a function... @ribhu please enlighten me as you seem to have good real analysis knowledge :)

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