#1

Ok, so I have a pre-calculus exam tomorrow, and I don't know how to solve inverse functions, which is going to be a big chunk on it. Someone please explain...

#2

Try this

#3

you see i would have never thought of asking a group of guitarists about maths.

i would have gone to a maths forum, kudos for thinking outside the box.

i would have gone to a maths forum, kudos for thinking outside the box.

#4

Haha tomorrow? Better lay on the caffeine I just finished school for the term so I can't remember how to do any of that crap

#5

#6

ok solving inverse functions like lets say Y= x^2 plus 2

really simple function.

instead of solving for y solve for x. and then switch the variables again.

so the inverse function would be Y=square root of(X+2)

steps:

1. Y=x^2-2

2.Y+2=x^2

3. square root of(y+2) = x

4. then switch variables and square root of(x+2)=Y

really simple function.

instead of solving for y solve for x. and then switch the variables again.

so the inverse function would be Y=square root of(X+2)

steps:

1. Y=x^2-2

2.Y+2=x^2

3. square root of(y+2) = x

4. then switch variables and square root of(x+2)=Y

#7

Get out of the Pit and studdy.

#8

1)to find the inverse of a function, make the x and y switch spots.

Y<SUP>2</sup>=(x-3)<sup>2</sup>-4

to

X<SUP>2</sup>=(y-3)<SUP>2</sup>-4

2) SOlve for y

X<SUP>2</sup>=(y-3)<SUP>2</sup>-4

x+4<SUP>2</sup>=(y-3)<SUP>2</sup>

_________________ _______________

v (x+4)<SUP>2</sup>=v(y-3)<SUP>2</sup

(x+4)=y-3

x+7=y

****the<SUP>2</sup> means squared****

hope that helps

Y<SUP>2</sup>=(x-3)<sup>2</sup>-4

to

X<SUP>2</sup>=(y-3)<SUP>2</sup>-4

2) SOlve for y

X<SUP>2</sup>=(y-3)<SUP>2</sup>-4

x+4<SUP>2</sup>=(y-3)<SUP>2</sup>

_________________ _______________

v (x+4)<SUP>2</sup>=v(y-3)<SUP>2</sup

(x+4)=y-3

x+7=y

****the<SUP>2</sup> means squared****

hope that helps