Proving two functions are inverses of each other

👉 Learn how to show that two functions are inverses. The composition of two functions is using one function as the argument (input) of another function. In simple terms composition of two functions is putting one function inside another function. The composition of two functions that are inverse to each other yields x.
To show whether two functions are inverse of each other, we composite the two functions. If the result is x, then the two functions are inverse of each other and they are not inverses otherwise.
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Organized Videos:
✅ Find the Inverse of a Function
• Find the Inverse of a ...
✅ Find the Inverse of a Function | Linear
• Find the Inverse of a ...
✅ Find the Inverse of a Function | Rational
• Find the Inverse of a ...
✅ Find the Inverse of a Function | Cubic
• Find the Inverse of a ...
✅ Find the Inverse of a Function | Quadratic
• Find the Inverse of a ...
✅ Find the Inverse of a Function | Cube Root
• Find the Inverse of a ...
✅ Find the Inverse of a Function | Square Root
• Find the Inverse of a ...
✅ Find the Inverse of a Function | Table
• Find the Inverse of a ...
✅ Find the Inverse of a Function | Graph
• Find the Inverse of a ...
✅ How to Prove Two Functions are Inverses
• How to Prove Two Funct...
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[@brianmclogan]

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[@omarpro8626]

@@brianmclogan sir, i think there is something wrong with your explanation, I see that there is no an inverse function for the square root function, I'll explain why, Let f(x)= (x-4)^½ Then the range of this function is from the several [0,infinity) Because its a square root You know that the domain of the inverse function = the range of the original function So therefore the inverse of that function is (x²+4) The range of that function is R. Thats just my explanation. I hope you reply to me ill appreciate that.

[@owenhildebrand6745]

I'm assuming you meant the domain of that function is R because you gave the range of the first one but im trying to figure this out and i don't see any error in your thinking because the domain of x²+4 is all real numbers because theres nothing you couldnt plug in for x and the range of (x-4)^½ would be zero to infinity because you cant get a negative from a square root. Those are supposed to be the same thing because the range of the original should be the domain of the inverse and vice versa but clearly you just found a situation where it isn't.

[@owenhildebrand6745]

okay i looked into this and i found the following rule," The square function is the inverse of the square root function. However the square root function is the inverse of the square function only if the domain of the square function is restricted to the positive numbers." Well if the domain is restricted to positive numbers it would be zero to infinity which matches up. I don't know why it works like that but it does i guess

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[@brianmclogan]

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Hi sir .... would you please tell me how to say two functions are invertible

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[@brianmclogan]

+Ashley Garner you are very welcome! happy to be able to assist you

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[@brianmclogan]

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do you have videos for finding either (fof)(x) or (gog)(x) when f(x)= square root of x squared plus 4. I can not find a video or any examples for either (fof)(x) or (gog)(x)...... THANKS!!!!

[@brianmclogan]

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[@brianmclogan]

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f(x)=square root x²-1 g(x) square root of x²+1

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So if i solve g(x) = x3(cube)- 4 is the answer still X?

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[@kerchcasenas662]

How to verify if two functions are inverse if the other function is a quadratic?

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[@brianmclogan]

Kevin Bhatt agree, happy to help!

[@imtherealtammygonna6216]

+Mr. McLogan's Math Channel What if you have a cubed root square root. like 3√4x-5

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Thanks this helps, but I had to watch it a few times because you go too fast.

[@brianmclogan]

yes sorry, It is not the first time, this is one of my earlier videos I have slowed it down recently

[@reincarn8tion]

I don't understand...g(x) isn't one-to-one so it can't have an inverse...right?

[@JacklyTranTheBeyMask]

So if you got something like -4x-20 and -4x-6 , they are not inverses?

[@robincsulb]

They are not inverses bc it's not a one-to-one function

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[@Tanzeel-qk9hc]

Another way we can prove this by making the graph of one and by taking the mirror on the line y= X and takes the reflection

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[@brianmclogan]

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[@racejohnson5658]

So if we had f(x) = -32x³ + 8 and g(x) = -(3׳ √-3x - 5), the answer is no, it's not a function. But if we had f(x) = √-1/2x + 8 and g(x) = -2(-x + 8)², x greater than or equal to 8, the answer is yes, it's a function.

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11/12/21 Precalculus senior high semester 1

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[@brianmclogan]

+CpSweetCable Good luck!

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[@andresgomez5220]

Wouldn't it be important to define limits on function f(x)? it doesn't have a value in real numbers before 4 and thus you wouldn't be able to cancel out the square root

[@fatihazehar]

Yes, the first thing we should do is finding the domain of each function

[@oleksandrmarkovichenko5774]

2:00 oof, sqrt(x^2) is not just "canceling out" to x. It is |x| in general, and it's "canceling out" to x only if you know that the domain is positive. You really should pay attention to this, especially when explaining it to the fresh students. That's why students don't understand why it is important to check that BOTH f(g(x))=x and g(f(x))=x. But looks like the guy from the video also don't understand why

[@Abeltech408]

you have a mistake on the square root of x square is not equlas to x make it corection please

[@yourmind.005]

What is condition of x. You took x is grater then zero

[@brianmclogan]

if they are inverses of each other it will not matter

[@chrischappa962]

Actually, the condition is necessary. Without it, you’d have things like f(g(-2)) = 2, violating the definition of being inverses. In the big picture, there is more than f(g(x)) = x and g(f(x)) = x. You must also have the domain of f(x) matching the range of g(x), and vice versa, the domain of g(x) matching the range of f(x). If there is a mismatch, then there exists values you can use as an input for one function, but not get as an output for the other function. Without the restriction written on the board, the mismatch is the domain of g(x) (all real numbers) doesn’t match the range of f(x) (all nonnegative real number). So any negative number could go into g(x), but not come back out of f(x).

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[@brianmclogan]

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[@fatihazehar]

F(g(x))=|x| because: √x²= |x|

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