## anonymous 4 years ago h(x)=6x-x^2 for x≥3 Express h−1(x) in terms of x.

1. anonymous

sorry thats express h inverse in terms of x

2. anonymous

Can't be done fuction is not bijective

3. anonymous

$x^2-6x=y$$x^2-6x+3=y+3$$(x-sqrt(3))^2=y+3$$x-\sqrt(3)=\pm(y+3)^2$$x=\pm(y+3)^2+sqrt(3)$

4. anonymous

Necessary and compulsory condition for inversion is that the function has to be bijective