Quick question --> inverse functions (1 Viewer)

[blackops23]

Hi guys, quick question, I'm kinda new to inverse functions so I need a bit of help..

We have f(x) = x^2 + 2x + 2 = (x+1)^2 + 1

Find the inverse function of f(x)

My working:

Ok so restrict the domain to x(>)-1

Then x = (y+1)^2 + 1
x-1 = (y+1)^2
therefore y = +/- (sqrt(x-1)) - 1

So as you can see there are two functions... but if I restricted the domain to x> -1 which function do I choose? Or do I have to restrict it to x>1?

Please explain guys, help immensely appreciate!

 

[Hermes1]

Hi guys, quick question, I'm kinda new to inverse functions so I need a bit of help..

We have f(x) = x^2 + 2x + 2 = (x+1)^2 + 1

Find the inverse function of f(x)

My working:

Ok so restrict the domain to x(>)-1

Then x = (y+1)^2 + 1
x-1 = (y+1)^2
therefore y = +/- (sqrt(x-1)) - 1

So as you can see there are two functions... but if I restricted the domain to x> -1 which function do I choose? Or do I have to restrict it to x>1?

Please explain guys, help immensely appreciate!

since the domain of f(x) = range of inverse of f(x)
therefore inverse of f(x) has range y > -1
hence u wuld take -1 + sqrt(x-1) as ur answer
(noting that the domain of inverse will be x > 1)

 

[Hermes1]

dude also make sure u no how to draw the inverse of a function (you reflect it about the line y = x)
also note the point of intersections of f(x) and its inverse will always meet on the line y=x
they sometimes try and trick u by asking you to find the POI of a really weird function, and if u dont let it equal to x you r goin to be spending all day trying to find its inverse.

 

[Hermes1]

Ahh ok thank you, another question:

Say you have the function y= x(inverse sinx) --> How would you find the domain and range? and how would you graph it?

Thanks for the help

first of all to graph it, you would hav to draw the graph of y = x and y = inverse sin x and multiply them together. I think the domain and range of the graph would be the domain and range of y = inverse sin of x.
ill explain this:
if you draw the graphs of y = x and y = inverse sin x

youll have inverse sin x between x = 1 and -1. at x = 1, y = x will give f(x) = 1 and y = inverse sin x will give f(x) = pi/2.

therefore multiplying them together gives you the value of y = xinverse sin x at x = 1 which is the extremity of the domain.

y = x and y = x sin inverse x multiplied together in the third quadrant produce a positive result. this is confirmed by the fact it is an even function. hence the domain is -1 to x to 1 inclusive and the range 0 to y to pi/2 inclusive

 

[Hermes1]

i had a bit of a fuk up b4. xsin inverse x is actually an even function. therefore range above should be 0 to y to pi/2 inclusive.
and domain stays the same as above.