Let f(x) = 2x - 6 Solve f-1(x) when x = 2.

Question

anonymous · Answer

-2

austinl · Answer

Do you mean, 

$f^{-1}(x)$

austinl · Answer

@sabrinasabor111 we don't just give answers here. http://openstudy.com/code-of-conduct

anonymous · Answer

sorry

anonymous · Answer

yes @austinL  thats what it is

austinl · Answer

Ok, we have:

$f(x) = 2x - 6$

Do find the inverse, we replace f(x) with y, and then solve for x. Once we have done that, we swap the x and y variables, then change y to f^-1(x).

$f(x) = 2x - 6$
$y=2x-6$
$2x=y+6$
$\displaystyle x=\frac{y+6}{2}$
$\displaystyle y=\frac{x+6}{2}$

$\displaystyle f^{-1}(x)=\frac{x+6}{2}$

Then to find the last part, you just input 2 for all instances of x!

$f^{-1}(2)=~?$

anonymous · Answer

4

austinl · Answer

Correctomundo!

anonymous · Answer

thanks can you help me with another one please?

austinl · Answer

Sure.

anonymous · Answer

Let f(x) = 2x + 2. Solve f-1(x) when x = 4.

austinl · Answer

Ok, how would you find the inverse of that f(x)?

Use the exact same steps as above :)

anonymous · Answer

f(x)= 2x+2
y=2x+2
2x= y+ 2
x= y+2
     ----
       2
y=x+2
   -----
      2

f^-1(x)=x+2
            -----
                2
             (4)+2
             -----
                 2

                 6
                 -
                 2

                 3 is the final answer

anonymous · Answer

@austinL did I get it?

austinl · Answer

Um, back at the top of your work. 

y=2x+2

What do you do to cancel out an addition?

anonymous · Answer

-2

austinl · Answer

Exactly, so we would have,

2x= y - 2

austinl · Answer

Can you continue from there?

anonymous · Answer

2x=y-2
x=y-2
    ----
      2
y=y-2
    ---
      2
    4-2
    ---
      2

      2
     --
      2
 so the answer is 1

austinl · Answer

I'd think so :)

austinl · Answer

Good job!

anonymous · Answer

okay I have one last question:
Let f(x) = x^2 - 16. Find f-1(x).
I don't know how to do it because of the exponent

austinl · Answer

To cancel out a square, you take the square root :) 
Make sense?

anonymous · Answer

a little care to show an example

anonymous · Answer

would it become  y= x + x - 16?

austinl · Answer

Lets say, 

$f(x)=x^2-25$

$y=x^2-25$

$y+25=x^2$

$x=\sqrt{y}+5$

Then swap etc :)

Now can you apply it to your problem?

anonymous · Answer

f(x) = X^2 -16
y = x^2 - 16
y + 16 = x^2
x= $$\sqrt{y+16}$$

anonymous · Answer

y= $$\sqrt{x+16}$$

austinl · Answer

$f^{-1}(x)=\sqrt{x}+4$

anonymous · Answer

okay thank you for all your help!

austinl · Answer

No problem, my pleasure!