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Find f^-1(x) and state the domain of f^-1 for

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f(x)=3x^2+5x-2. , x>=0?

asked Jun 5, 2013 in ALGEBRA 1 by futai Scholar

1 Answer

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f (x )  = 3x^2 + 5x -2 ;   x> = 0

y = 3x^2 + 5x - 2

3x^2 + 5x - 2 - y  = 0

3x^2 + 5x - ( 2 + y ) = 0

 Therefore  x  = {- b ± (√(b^2 - 4ac))} / 2a

                       = {-5 ±  {√(25 - 4 (3) [ - ( 2+y )]}} / 6

                       = { -5 ±  √(25 + 24 + 12y)} / 6

                      =  {-5 ±  √(49 + 12y)} / 6

                     ={ -5±  √( 49 + 12y)} / 6         ( x > = 0 )

        Replace  y = x

   y = { -5 ±  √( 49 + 12x)} / 6

f'(x) = { -5 ±  √( 49 + 12x)} / 6

    49 + 12X  =  5    (√(49 + 12X) not equal  to  5)

Take square each side

  49 + 12 x  =  25

 12x  25 - 49

 Therefore  x = - 24 / 12 = - 2

 Domain = x  belongs  to ( - infinity , infinity ) - { - 2 }.

answered Jun 11, 2013 by goushi Pupil

The inverse function f(x) = (1/6)[- 5 ± √(12x + 49)].

Domain of a function f(x) is set of those values of x which will make the function mathematically legal or correct..certain operations like division by zero , square root of a negative number do not exist in real maths.

So they are prohibited .so we have to look carefully for such possibilities to determine the domain of the function..here we have one operations present.

So, the domain of the above function is all non negative real numbers.

√(12x + 49) ≥ 0

12x + 49 ≥ 0

12x ≥ - 49

x ≥ - 49/12.

The domain of inverse function f(x) = (1/6)[- 5 ± √(12x + 49)] is {x ∈ R : x ≥ - 49/12}.

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