i used to do these in my sleep.. But not using it in my field makes one forget all this easy stuff. Now i know why my dad struggled with the math i did.. lol he was a Masters level engineer.
sir on the third step you can directly do the tp process I mean just simply multiply the denominator to the numerator on the other hand side why didn't you do so?
Substitute y for f(x), swap x and y, then solve for y. For example f(x) = x^2 - 4x + 4 y = x^2 - 4x + 2 x = (y - 2)^2 y - 2 = x^(1/2) y = 2 + x^(1/2) You can also check since f[f^-1(x)] = x = f^-1[f(x)]. In the case of my example, however, we need to be careful since we have even powers and even roots. If you check the negative x values, you see a problem with the inverse that means an unrestricted domain excludes the inverse from being a function. In order to be all encompassing, we would actually need two inverse functions with restricted domains and a sign change.
The notation signifies the inverse of function f is also a function. Since f^-1(7) outputs 3, the definition of a function means _only_ f(3) can output 7.
Even if it has a multivalued inverse, this method still produces at least one valid value of t. To conclude that it is the only valid value of t, you'd have to also know that f(x) is a 1-to-1 function.
The name of an unspecified function. It is assumed for lack of other information, that f(x) is a one-to-one function, with only a single-valued inverse.
Bruh that chalking tapping while doing math is on another level.
He said he has been doing that for 25 years. 😊
I love it when you look at the camera instead of looking what you wrote at the end
you should really state the your method depends on the invertability of f otherwise we could not be sure that f(x)=f(y) implies x=y
I have a regular video explaining that.
It's not possible to explain everything in UA-cam shorts.
Nice way of doing it. It's always necessary to remember algoritms! 👍
awesome!
only works with one to one functions otherwise we are not sure if there is only one unique x whose output is 7
I am not even understanding what he's saying but i am understanding his doing maths 😂
i used to do these in my sleep.. But not using it in my field makes one forget all this easy stuff. Now i know why my dad struggled with the math i did.. lol he was a Masters level engineer.
You’re awesome ❤
Lol
sir
on the third step you can directly do the tp process
I mean just simply multiply the denominator to the numerator on the other hand side
why didn't you do so?
Because in some fractions you have two denominators and it's better to get use with that method. 👍
@@luisclementeortegasegovia8603 okay that's cool but
The method which is I am discussing is also applicable for the problem you discussed......
Just for an example
1 + 2t/ (3 + t)(2 + 7t) = 5
1st method
1 + 2t = 5 (3 + t)(2 + 7t)
1 + 2t = 5 (6 + 21t + 2t + 7t²)
1 + 2t = 30 + (5 x 23t) + 35t²
1 + 2t = 30 + 115t + 35t²
.....continued
2nd method
1 + 2t/ (3 + t)(2 + 7t) = 5
1 + 2t/ (6 + 21t + 2t + 7t²) = 5
1 + 2t = 5 (6 + 21t + 2t + 7t²)
1 + 2t = 30 + (5 x 23t) + 35t²
1 + 2t = 30 + 115t + 35t²
.....continued
we casually use the 2nd method.......
can u elaborate how inverse of a function work?
Substitute y for f(x), swap x and y, then solve for y. For example
f(x) = x^2 - 4x + 4
y = x^2 - 4x + 2
x = (y - 2)^2
y - 2 = x^(1/2)
y = 2 + x^(1/2)
You can also check since f[f^-1(x)] = x = f^-1[f(x)].
In the case of my example, however, we need to be careful since we have even powers and even roots. If you check the negative x values, you see a problem with the inverse that means an unrestricted domain excludes the inverse from being a function. In order to be all encompassing, we would actually need two inverse functions with restricted domains and a sign change.
I don’t think you find enough solution for the problem, because there might be different values of a and b such that f(a) = f(b)
How can we set them equal as many functions have same value for different values of x can someone explain pls or this is a special case
The notation signifies the inverse of function f is also a function. Since f^-1(7) outputs 3, the definition of a function means _only_ f(3) can output 7.
Even if it has a multivalued inverse, this method still produces at least one valid value of t.
To conclude that it is the only valid value of t, you'd have to also know that f(x) is a 1-to-1 function.
Is it assumed that f is one-one?
Yes. It is required that the f(x) is a one-to-one function for this reasoning to be valid.
If a function has an inverse, it means that they are bijective (each y has one and only one x)
Please the answer is just -4
=> 1/f(7)=3 f(7)=1/3 f[(1-2t)/(1+2t)]=f(7) f[(1-2i)/(1+2i)]=-i/3i=-1/3; f[(1-2•1)/(1+2•1)]=-1/3 t1=| i |; t2= |1|
What is f
The name of an unspecified function. It is assumed for lack of other information, that f(x) is a one-to-one function, with only a single-valued inverse.
imagine assuming the inverse exists without proving it
Imagine assuming that f((1-2t)/(1+2t)) is even defined. That's just the data given in the exercise, they are not gonna give you false information.
@@bambouejfr9263 you get 0 marks for making any such assumption of existence. Not sure what brain dead take you are on.