Algebra (all content)
- Finding inverse functions: linear
- Finding inverse functions: quadratic
- Finding inverse functions: quadratic (example 2)
- Finding inverse functions: radical
- Finding inverses of rational functions
- Finding inverse functions
- Finding inverses of linear functions
Finding inverse functions: quadratic (example 2)
Sal finds the inverse of f(x)=(x-1)^2-2. Created by Sal Khan.
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- @4:22-4:41.... I do not understand why is the neg. sign taken out from the square root?
How does this -√(x-1)^2 become only x-1 ?(148 votes)
- I think the easiest way to understand this is to use the process of solving a quadratic by completing the square as a reference. For example, let's say you complete the square on a quadratic and get:
(x + 8)^2 = 121
When you take the square root of both sides you end up with:
x + 8 = +/-11
Note that the square root of (x + 8)^2 is just x + 8, but that it is EQUAL to positive 11 or negative 11; this equality is explicitly stating that the square root of (x + 8)^2 can have both a positive or negative value. Now, if we look at the problem in this video:
(x - 1)^2 = y + 2
See how its format looks identical to the completing the square example? But, in this case we know that because the domain is restricted to x<=1, that when we take the square root of each side we're not concerned with both the positive and negative square root of y + 2; we only want the negative square root of y + 2. So, when we take the square root of each side, we end up with:
x -1 = -√(y + 2)(20 votes)
- at2:45it says it's negative, what about if x is 1?(21 votes)
- by negative he actually means non positive which also includes zero. Hope this helps!!!(32 votes)
- Why are function and its inverse symmetrical across the line y=x? How to generally prove that?(18 votes)
- 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(9 votes)
- Is the inverse of a function created when you just switch the
yin an equation?(10 votes)
- 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).(13 votes)
- 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
sqrt((x-1)^2) = x-1
-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?(8 votes)
- 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)
sqrt((x-1)^2) = -x-1
if x= -3
then (-3-1)^2 = (-4(15 votes)
- At4:57how 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.(8 votes)
- 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 <= 1pulled along to every step, to remember which side to use.(6 votes)
- 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(4 votes)
- 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 :)(2 votes)
- 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?(3 votes)
- The inverse function of y, (which is our answer at6:20), wouldn't be the inverse function of x or y of our original equation, right? Because all we did to our original equation is solve for x in terms of y instead of y in terms of x. In order for it to be the inverse, we would HAVE to make it a function of x, correct?(4 votes)
- Are there functions that don't have an inverse function ? For instance, a function which associates a same 'y' to several different 'x' values ; logically its inverse would have to associate several range values to a same domain value and thus not be a function, wouldn't it ?(3 votes)
- Yes, you're completely right. One such function is y=x^2, whose inverse is not a function (y=sqrt(x)).(2 votes)
We have the function f of x is equal to x minus 1 squared minus 2. And they've constrained the domain to x being less than or equal to 1. So we have the left half of a parabola right here. They've constrained so that it's not a full U parabola. And I'll let you think about why that would make finding the inverse difficult. But let's try to find the inverse here. And a good place to start -- let's just set y being equal to f of x. You could say y is equal to f of x or we could just write that y is equal to x minus 1 squared minus 2. We know it's for x is less than or equal to 1. But right now we have y solved for in terms of x. Or we've solved for y, to find the inverse we're going to want to solve for x in terms of y. And we're going to constrain y similarly. We could look at the graph and we could say, well, in this graph right here, this is defined for y being greater than or equal to negative 2. So we can maybe put in parentheses y being greater than or equal to negative 2. Because this is going to be -- right now this is our range. But when we swap the x's and y's, this is going to be our domain. So let's just keep that in parentheses right there. So let's solve for x. So that's all you have to do, to find the inverse. Solve for x and make sure you keep track of the domains and the ranges. So, let's see. We could add 2 to both sides of this equation. We get y plus 2 is equal to x minus 1 squared. Minus 2, plus 2, so those cancel out. And then, I'm just going to switch to the y constraint. Because now it's not clear what we're -- whether x is the domain or the range. But we know by the end of this problem, y is going to be the domain. So let's just swap this here. So, 4 for y is greater than or equal to negative 2. And we could also say, in parentheses, x is less than 1. This is -- we haven't solved it explicitly for either one, so we'll keep both of them around right now. Now, to solve for x, you might be tempted to just take the square root of both sides here. And you wouldn't be completely wrong. But we have to be very, very, very careful here. And this might not be something that you've ever seen before. So this is an interesting point here. We want the right side to just be x minus 1. That's our goal here, in taking the square root of both sides. We want to just have an x minus 1 over there. Now, is x minus 1 a positive or a negative number? Well, we've constrained our x's to being less than 1. So we're dealing only in a situation where x is less than or equal to 1. So if x is less than or equal to 1, this is negative. This is negative. So we want to take the negative square root. Let me just be very clear here. If I take negative 3 -- I take negative 3 and I were to square it, that is equal to 9. Now, if we take the square root of 9 -- if we take the square root -- let's say we take the square root of both sides of this equation. And our goal is to get back to negative 3. If we take the positive square root. If we just take the principle root of both sides of that, we would get 3 is equal to 3. But that's not our goal. We want to get back to negative 3. So we want to take the negative square root of our square. So because this expression is negative and we want to get back to this expression, we want to get back to this x minus 1, we need to take the negative square root of both sides. You can always -- every perfect square has a positive or a negative root. The principle root is a positive root. But here we want to take the negative root because this expression right here is going to be negative. And that's what we want to solve for. So let's take the negative root of both sides. So you get the negative square root of y plus 2 is equal to -- and I'll just write this extra step here, just so you realize what we're doing. Is equal to the negative square root, the negative square root of x minus 1 squared. For y is greater than or equal to negative 2. And x is less than or equal to 1. That's why the whole reason we're going to take the negative square there. And then this expression right here -- so let me just write the left again. Negative square root of y plus 2 is equal to the negative square root of x minus 1 squared is just going to be x minus 1. It's just going to be x minus one. x minus 1 squared is some positive quantity. The negative root is the negative number that you have to square to get it. To get x minus 1 squared. So that just becomes x minus 1. Hopefully that doesn't confuse you too much. We just want to get rid of this squared sign. We want to make sure we get the negative version. We don't want the positive version, which would have been 1 minus x. Don't want to confuse you. So here, we now just have to solve for x. Add 1. And let me write the 4. y is greater than or equal to negative 2. Add 1 to both sides. You get negative square root of y plus 2 plus 1 is equal to x for y is greater than or equal to negative 2. Or, if we want to rewrite it, we could say that x is equal to the negative square root of y plus 2 plus 1 for y is greater than or equal to negative 2. Or if we want to write it in terms, as an inverse function of y, we could say -- so we could say that f inverse of y is equal to this, or f inverse of y is equal to the negative square root of y plus 2 plus 1, for y is greater than or equal to negative 2. And now, if we wanted this in terms of x. If we just want to rename y as x we just replace the y's with x's. So we could write f inverse of x -- I'm just renaming the y here. Is equal to the negative square root of x plus 2 plus 1 for, I'm just renaming the y, for x is greater than or equal to negative 2. And if we were to graph this, let's see. If we started at x is equal to negative 2, this is 0. So the point negative 2, 1 is going to be on our graph. So negative 2, 1 is going to be on our graph. Let's see, if we go to negative 1, negative 1 this will become a negative 1. Negative 1 is 0 on our graph. Negative 1, 0 is on our graph. And then, let's see. If we were to do, if we were to put x is equal to 2 here. So x is equal to 2 is 4. 4 square root, principle root is 2. It becomes a negative 2 So it becomes 2, negative 1. So that's on our graph right there. So the graph is going to look something like this, of f inverse. It's going to look something like that right there. As you can see, it is a reflection of our original f of x along the line y is equal to x. Along the line y is equal to x. Because we've essentially just swapped the x and the y. This is about as hard of an inverse problem that I expect you to see. Especially in a precalculus class because it really is tricky to realize that you have to take the negative square root here. Because the way our domain was constrained, this value right here is going to be negative. So to solve for it, you want to have the negative square root.