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mukushla

  • 3 years ago

Range of\[f(x)=\sqrt[3]{x-1}+\sqrt[3]{3-x}\]

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  1. mukushla
    • 3 years ago
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    ????

  2. akitav
    • 3 years ago
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    made a stupid error, just a sec.

  3. akitav
    • 3 years ago
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    |dw:1347772271638:dw|

  4. mukushla
    • 3 years ago
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    yeah range is limited...i dont how to evaluate the exact interval for range :(

  5. mukushla
    • 3 years ago
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    *know

  6. akitav
    • 3 years ago
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    Can you plot the two parts of the function? then combine them.|dw:1347772792991:dw| For analyzing the final result : If it's a diagonal line then yes both will go from -infinity to +infinity, IF HOWEVER it is a VERTICAL or HORIZONTAL line then your domain or range will be limited.

  7. mukushla
    • 3 years ago
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    i want to show it algebrically if its possible !!

  8. artofspeed
    • 3 years ago
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    idea: seperate f(x) into 2 functions, find the domain of the inverse of each one of the function, find the intersection of the 2 domains, that gives the range of f(x)

  9. nipunmalhotra93
    • 3 years ago
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    Range is ]0,2]. f(x) is symmetrical about x=2. I evaluated the first and second derivatives. for x<1<2 f'(x)>0 and f''(x)>0 for x<1, f'(x)>0 and f''(x)<0. behavior on other side of 2 is symmetrical. Also, evaluating the limit at -infinity, we get 0. So the graph would be something like this.|dw:1347780606499:dw| That's it. Correct me if I'm wrong.

  10. nipunmalhotra93
    • 3 years ago
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    Another small test is that f(x) can't be <0 because solving it gives -1<-3 which is a contradiction.

  11. harsh314
    • 3 years ago
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    the range of f(x) is ( 0,2]

  12. harsh314
    • 3 years ago
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    to show it algebrically you need to assume \[\sqrt[3]{x-1}= \alpha\] \[\sqrt[3]{3-x}= \beta\] and after that you can obtain the required equations

  13. harsh314
    • 3 years ago
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    as \[f(x)=y=\alpha +\beta\] so find out x in form of y and you would get the answer if you want more detailed solution then i can provide you that too

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