INVERSE FUCTIONS....MATH HELP.....10 PTS.....///read///?

Question

Says:

Determine the inverse functions for the functions below. Begin by switching varaiables in the original function, then solving for the 'new' y. For each case sketch the function and its inverse.

Have no idea what this is or what to do. Please explain how u found answer. 10 Points! :)

3. y= (-1/2)x + 7

What is the inverse fuction for this and what in the invariant?

Answer 1

It might be easier to understand in this format:

f(x) = (-1/2)*x + 7

To determine its inverse, treat f(x) and x as variables, then switch them, then solve for the new f(x) [we'll call it f'(x)]:

x = (-1/2) * f'(x) + 7

x - 7 = (-1/2) * f'(x)

f'(x) = -2 * (x - 7)

f'(x) = -2*x + 14

Answer 2

y = -(1/2)x + 7

x = -(1/2)y + 7 --switch variables

-x = 1/2y - 7 --multiply both sides by -1

7-x = 1/2y --add 7 to both sides

14-2x = y --multiply both sides by 2

f'(x) = 14-2x

Basically, the idea is that you have x, y and f(x)=y, which is a mapping from x to y.  X is the independant variable, y is the dependant variable.

The inverse function is a mapping from y to x, so if you take the function f(x) and want the inverse function f'(y) = x, then you take the "body" of f(x), swap x for y giving you the function f(y) = x which is _not equivalent_ to f'(y)=x because the body doesn't solve for x.  You want an equality which states the dependant variable on it's own as equalling some function of "x") so you re-arrange for the dependant variable, y.

et voila

Answer 3

Do exactly what the instructions say. Where there is a "y" in the given equation, replace it with "x"; where there is an "x" in the given equation, replace it with "y"

x = (-1/2)y + 7

Now solve for y:

x - 7 = (-1/2)y

(x - 7)/(-1/2) = y

-2(x - 7) = y

-2x + 14 = y

I don't know what "the invariant" means in this context. If I had to guess, I would say that it is the value of x that gives the same value of y for the given function and its inverse. If so,

(-1/2)x + 7 = -2x + 14

(3/2)x = 7

x = 14/3

Answer 4

Here's a good explanation:

http://www.mathclues.com/ModuleFindAnInv...

Hope this helps you!  Please email anytime for further questions.