inverse functions/trig (1 Viewer)

[.ben]

Find th einverse of the the following:

1. y=(x-2)² and x=<1

just interchanging x and y i know but do you have to show the domain of the inverse or not. in some solutions they replace all x's with y's including the domain of the original function so that the solution is: y=1-sqrt(x) y=<1. is that right or do you go x>0 for the domain?

2. a) find d/dx(sin^-1x+cos^-1x)
b) hence prove sin^-1x+cos^-1x=pi/2

is it possible to integrate part a) both sides and obtain a C for constant so that it can be any value?

e.g. int(d/dx(sin^-1x+cos^-1x)dx=int(0)dx
sin^-1x+cos^-1x=C

 

[Slidey]

It is possible to integrate part a and get a constant... but it turns out that constant will always be pi/2.

 

[Riviet]

Find th einverse of the the following:

1. y=(x-2)² and x=<1

just interchanging x and y i know but do you have to show the domain of the inverse or not. in some solutions they replace all x's with y's including the domain of the original function so that the solution is: y=1-sqrt(x) y=<1. is that right or do you go x>0 for the domain?

You need to see what the question is asking for, for a parabola, it's inverse will NOT be a function because there are two values of y for every x value when you look at the inverse. For your example, the question has specified x<, which is the left half of the parabola. Therefore, when you find the inverse, you only take the half the parabola which corresponds to the inse of the half parabola in the original function. So your answer is x=2-sqrt(y) which is the bottom half of the parabola x=2+sqrt(y)