Math P1 s07

[mz]

can anybody please explain the concept behind question number 11, part ii?

[immortal]

Find da inverse of f(x) dat is f^-1(x)
Domain of f^-1(x)=range of f(x) & vice-versa

[nid404]

can anybody please explain the concept behind question number 11, part ii?

here u go


Let f-1(x) = y
i.e. f(y) = x

x= 6/ 2y+3

2y+3=6/x
2y= 6/x -3
y= (6-3x)/2

thus f^-1(x)= (6-3x)/2
y >= 0 implies that 6-3x >= 0 i.e.
6 >= 3x  i.e.  x <= 2
Thus the domain of f-1(x) is 0 <=  x  <= 2

[mz]

thankyou so much both of you!!!!