Find the product of maximum and minimum value of $f(x)=sqrt{2}sin{x}-f(0)$












1












$begingroup$


$f(x)=sqrt{2}sin{x}-f(0)$



(A) $frac{-3sqrt{2}}{4}$



(B) $frac{-3sqrt{2}}{2}$



(C) $frac{-3}{2}$



(D) $frac{-3}{4}$



(E) $frac{-7}{4}$



My attempt :



let x=0



$f(0)=sqrt{2}sin{0}-f(0)\
f(0)=-f(0)\
2f(0)=0\
f(0)=0$



then :



$f(x)=sqrt{2}sin{x}$



the answer may be $-sqrt{2}$ but it's not in the option.






calculus differential







share|cite|improve this question









$endgroup$












  • $begingroup$
    The answer is $(sqrt 2) (-sqrt 2)=-2$ which is not one of the options.
    $endgroup$
    – Kavi Rama Murthy
    Jan 22 at 12:19
















1












$begingroup$


$f(x)=sqrt{2}sin{x}-f(0)$



(A) $frac{-3sqrt{2}}{4}$



(B) $frac{-3sqrt{2}}{2}$



(C) $frac{-3}{2}$



(D) $frac{-3}{4}$



(E) $frac{-7}{4}$



My attempt :



let x=0



$f(0)=sqrt{2}sin{0}-f(0)\
f(0)=-f(0)\
2f(0)=0\
f(0)=0$



then :



$f(x)=sqrt{2}sin{x}$



the answer may be $-sqrt{2}$ but it's not in the option.






calculus differential







share|cite|improve this question









$endgroup$












  • $begingroup$
    The answer is $(sqrt 2) (-sqrt 2)=-2$ which is not one of the options.
    $endgroup$
    – Kavi Rama Murthy
    Jan 22 at 12:19














1












1








1





$begingroup$


$f(x)=sqrt{2}sin{x}-f(0)$



(A) $frac{-3sqrt{2}}{4}$



(B) $frac{-3sqrt{2}}{2}$



(C) $frac{-3}{2}$



(D) $frac{-3}{4}$



(E) $frac{-7}{4}$



My attempt :



let x=0



$f(0)=sqrt{2}sin{0}-f(0)\
f(0)=-f(0)\
2f(0)=0\
f(0)=0$



then :



$f(x)=sqrt{2}sin{x}$



the answer may be $-sqrt{2}$ but it's not in the option.






calculus differential







share|cite|improve this question









$endgroup$




$f(x)=sqrt{2}sin{x}-f(0)$



(A) $frac{-3sqrt{2}}{4}$



(B) $frac{-3sqrt{2}}{2}$



(C) $frac{-3}{2}$



(D) $frac{-3}{4}$



(E) $frac{-7}{4}$



My attempt :



let x=0



$f(0)=sqrt{2}sin{0}-f(0)\
f(0)=-f(0)\
2f(0)=0\
f(0)=0$



then :



$f(x)=sqrt{2}sin{x}$



the answer may be $-sqrt{2}$ but it's not in the option.






calculus differential




calculus differential






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 22 at 12:10









airlanggaairlangga

775




775












  • $begingroup$
    The answer is $(sqrt 2) (-sqrt 2)=-2$ which is not one of the options.
    $endgroup$
    – Kavi Rama Murthy
    Jan 22 at 12:19


















  • $begingroup$
    The answer is $(sqrt 2) (-sqrt 2)=-2$ which is not one of the options.
    $endgroup$
    – Kavi Rama Murthy
    Jan 22 at 12:19
















$begingroup$
The answer is $(sqrt 2) (-sqrt 2)=-2$ which is not one of the options.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 12:19




$begingroup$
The answer is $(sqrt 2) (-sqrt 2)=-2$ which is not one of the options.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 12:19










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$begingroup$

You've done the tough bit!



Now remember that $-1leq sin xleq 1$, so what would be the bounds of $sqrt2 sin x$?



As it turns out, none of the options you are given are correct.






share|cite|improve this answer









$endgroup$













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    1 Answer
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    1 Answer
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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    You've done the tough bit!



    Now remember that $-1leq sin xleq 1$, so what would be the bounds of $sqrt2 sin x$?



    As it turns out, none of the options you are given are correct.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You've done the tough bit!



      Now remember that $-1leq sin xleq 1$, so what would be the bounds of $sqrt2 sin x$?



      As it turns out, none of the options you are given are correct.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You've done the tough bit!



        Now remember that $-1leq sin xleq 1$, so what would be the bounds of $sqrt2 sin x$?



        As it turns out, none of the options you are given are correct.






        share|cite|improve this answer









        $endgroup$



        You've done the tough bit!



        Now remember that $-1leq sin xleq 1$, so what would be the bounds of $sqrt2 sin x$?



        As it turns out, none of the options you are given are correct.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 at 12:23









        Rhys HughesRhys Hughes

        6,9741530




        6,9741530






























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