If f(x)= 8+2x^2 + tan(Pi(x) \ 2), where -1<x<1, find f^-1(8)?

Question

This is an inverse function in a beginnning college calculus class, not too hard just dont know how to find the inverse of this particular problem. <<<<>>>>nothing just to get points

Answer 1

Let f^-1(x)=y, the inverse function y is definite by this equation:
x=8+2y^2 + tan(Pi(y) \ 2),
in the point x=8 for y(8) you have:
8=8+2y^2 + tan(Pi(y) \ 2) ->
2y^2 + tan(Pi(y) \ 2)=0 ->
y=0
Thus: f^-1(8)=0

Answer 2

well f(x1) = 8 +2x^2 so f^-1 (x) = f(y) = 8+ 2y^2
& f(x2) = tan(Pi(x) \ 2 so f^-1(x2) or Arctan (x) = tan (∏y)/ 2 .

f^-1(x) Or f(y) = 8+ 2y^2 + tan (∏y)/ 2

{ rememebr if we have tan(x) so Domain = ( - ∏/2 , + ∏/2)
& R = ( -∞,+ ∞) .But if we have y = arctan(x) <=> x = tan(y) , - ∏/2 < y < + ∏/2 }

f^-1(8) = 8+ 2(y)^2 + tan (∏(y))/ 2
{ as f^-1(x) = y and x = 8 so ;}
8 = 8+ 2(y)^2 + tan (∏(y))/ 2
8+ 2(y)^2 + tan (∏(y))/ 2 = 8
2(y)^2 + tan (∏(y))/ 2 = 8-8
2y^2 + tan (∏(y))/ 2 = 0
factor ' y '
y ( 2y + tan(∏/2) ) = 0
now
y = 0
Or
2y + tan(∏/2) = 0; y = tan(∏/2) / 2 ;y = tan∏/ (2*2) ; y = tan∏/4
y = +1 .

so its our correct answer we are looking for.
Good luck and Good question.

Answer 3

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