Question

given f(x)=sinx and g(x)=x^2+1, what is f(g(x)) and g(f(x))?

Answer 1

\[\Large f(x) = \sin(x)\]
\[\Large f(\color{red}{x}) = \sin(\color{red}{x})\]
\[\Large f(\color{red}{g(x)}) = \sin(\color{red}{g(x)})\]
Do you see how I replaced every x with g(x)?

Answer 2

that means you would end up with f(g(x))=sin^2x + 1 and using that, g(f(x)) = sin^2x+1 right?

Answer 3

well we replace the g(x) on the right side with x^2 + 1

Answer 4

so let me write out the full steps

\[\Large f(x) = \sin(x)\]
\[\Large f(\color{red}{x}) = \sin(\color{red}{x})\]
\[\Large f(\color{red}{g(x)}) = \sin(\color{red}{g(x)})\]
\[\Large f(\color{red}{g(x)}) = \sin\left(\color{red}{x^2+1}\right)\]

Answer 5

As for the other way around, it looks like you got it
\[\Large g(x) = x^2 + 1\]
\[\Large g(f(x)) = (f(x))^2 + 1\]
\[\Large g(f(x)) = (\sin(x))^2 + 1\]
\[\Large g(f(x)) = \sin^2(x) + 1\]

Answer 6

no you cannot distribute like that

Answer 7

`sin(x+1)` does NOT turn into `sin(x) + sin(1)`

Answer 8

wait, so, sin(x^2 +1) cannot be simplified anymore?

Answer 9

yeah it's as simplified as it gets