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### Course: Algebra (all content)>Unit 7

Lesson 20: Finding inverse functions (Algebra 2 level)

# Finding inverse functions: quadratic (example 2)

Sal finds the inverse of f(x)=(x-1)^2-2. Created by Sal Khan.

## Want to join the conversation?

• at it says it's negative, what about if x is 1?
• by negative he actually means non positive which also includes zero. Hope this helps!!!
• Why are function and its inverse symmetrical across the line y=x? How to generally prove that?
• The inverse of a function is the expression that you get when you solve for x (changing the y in the solution into x, and the isolated x into f(x), or y).

Because of that, for every point [x, y] in the original function, the point [y, x] will be on the inverse.

Let's find the point between those two points. Using the formula for the middle-point we get the following.
p[x] = (x + y) / 2
p[y] = (y + x) / 2
From that, p = [0.5 * (x + y), 0.5 * (x + y)]

Because the x and y values of p are equal, it lies on the line y = x.

I hope this helps.

--Phi φ
• Is the inverse of a function created when you just switch the `x` and `y` in an equation?
• Pretty much yes, but you have to be careful as to what you exactly mean by that. Just keep in mind that the inverse of a function is another function that has the output of the original function as its input, and the input of the original function as its output. That is, if the original function made you go from 2 (its input, which is x) to 4 (its output, which is y), then the inverse of this function would make you go from 4 (its input, which is x, and which was y in the original function) to 2 (its output, which is y, and which was x in the original function).
• I trust that the answer is correct but I still don't trust the explanation that

-sqrt((x-1)^2) = x-1

Here is my thinking...

-sqrt((x-1)^2) = x-1
and
sqrt((x-1)^2) = x-1
Therefore
-sqrt((x-1)^2) = sqrt((x-1)^2) I do not trust this to be true.

I solved the problem of the finding inverse f(x) = (x-1)^2-2 by substitution. Here is my thinking...

Substitute x and y. Solve for y...
x = (y-1)^2 - 2
x - 2 = (y-1)^2
sqrt(x-2) = y-1
-sqrt(x-2) = y
y = -sqrt(x-2)

Is my logic wrong?
• since in the beginning of the problem we limited x to x<=1 therefore (x-1) is a negative number

and when we square a negative we get a positive number (for ex. -2^2=4)

so sqrt((x-1)^2)would give us the positive version of (x-1) which is -(x-1)(since (x-1)is a negative number as we said before and negative a negative number gives us a positive for ex. -(-2) is 2)

so -sqrt((x-1)^2) = -(-(x-1))=x=1
and sqrt((x-1)^2) = -x-1 (not x-1)

for ex.

sqrt((x-1)^2) = -x-1

if x= -3
then (-3-1)^2 = (-4
• At how do you know when to take the negative square root or positive one? I didn't get how in the video. Also if you could tell me, is this level of math algebra 1 or 2? They should do a better job of subdividing the algebra category between the 1 and 2.
• Whether you take the positive or negative root depends on which side of the graph is present. That's why Sal kept the `for y >= -2 and x <= 1` pulled along to every step, to remember which side to use.
• Why is it reflected across the y=x line? I know it is for all inverse functions, but is there a reason for it? Thanks
• To add on to what Taha Junejo said, it is because you're basically switching the x and y-values. So basically, you're graphing the same function, except the y-axis becomes the x-axis and the x-axis becomes the y-axis. Therefore, it is reflected across the y=x line. I hope the clears your doubt :)
• How would you find the inverse of a function that is a fraction? Such as f(x)=6/(√x) ?
When I tried it, I got f^-1(x)=x^2-36, is that right?