If I need to get a lot of solutions of this form quickly, I might use the first method. But, for displaying the solution to others, or to myself to check my own work, I'd use the second method. They're pretty much equivalent but have different levels of explanatory power. You got to the solution faster with the first method of course, but with the second method I fully understood what you were doing whether I had previous learning on this or not.
Through the years I've often stumbled on this very simple concept, partly because of notation. Effectively we're drawing a diagonal line on the 2D graph plane, and reflecting the function by folding on that 45 degree line. So, if f(x) is e^x, then f-1(x) is ln(x). And if y=xe^(x), then the inverse is W(x). You give me y, I'll give you x, it's all about the input data.
I'll admit it always confuses me that we say both f(x) and f^-1(x), I find it easier to consider f(x) and f^-1(y), or at least use two different variables and solve for the one we care about in both cases; perhaps it's because I'm more on the physics side of things, where units matter and we're used to solve for things that aren't called x, but I find it easy to mess up and confuse which x it is in which context otherwise.
If I need to get a lot of solutions of this form quickly, I might use the first method. But, for displaying the solution to others, or to myself to check my own work, I'd use the second method. They're pretty much equivalent but have different levels of explanatory power. You got to the solution faster with the first method of course, but with the second method I fully understood what you were doing whether I had previous learning on this or not.
I agree, and we find the equation of the inverse function too.
Excellent video as always! 👍
If f(x)=3+(2x+1)^1/2, then
f(x-(2x+1)^1/2)=3 for this
f(4)=3+[x-(2x+1)^1/2]÷3, then
f^-1(4)=1/f^1(4)=3/3+[x-(2x+1)^1/2]
x=7
Through the years I've often stumbled on this very simple concept, partly because of notation. Effectively we're drawing a diagonal line on the 2D graph plane, and reflecting the function by folding on that 45 degree line. So, if f(x) is e^x, then f-1(x) is ln(x). And if y=xe^(x), then the inverse is W(x). You give me y, I'll give you x, it's all about the input data.
Thank you so much
Lots of love from India
I think first method is easier more than second one. Thank u teacher
Thank you
The first method is the easiest and fastest, while the second is the most complete, 'cause we find the equation of the inverse function.
y가 4일때 x는 얼마냐를 묻는 문제가 역함수 문제였군요..
I'll admit it always confuses me that we say both f(x) and f^-1(x), I find it easier to consider f(x) and f^-1(y), or at least use two different variables and solve for the one we care about in both cases; perhaps it's because I'm more on the physics side of things, where units matter and we're used to solve for things that aren't called x, but I find it easy to mess up and confuse which x it is in which context otherwise.
Thank you, some book report only one method ✌🏻✌🏻
Stop giving easy algebra questions. You're making me feel like i'm good at algebra
I like the second method!!!
The first method
🔥🔥🔥🔥🔥
0
Ez.
Done in mind in 5 seconds!
Proof math has nothing to do with intelligence..
Wow
Thank you