Solve the composition $f \circ g=?$ and $g\circ f=?$

Question

a) $f(x) = \sqrt[3]{x}\ $ and $g(x) = x^3$

find $f\circ g=?$ and $g\circ f=?$

I have $f\circ g = f(g(x)) = f(x^3) = \sqrt[3]{x^3} = x$

So basically, first we replace $g(x)$ with its value, then we replace $x$ with the value of $f(x)$ inside $g(x)$?

$g\circ f = g(f(x)) = g( \sqrt[3]{x} ) = \sqrt[3]{x^3} $ OR $ \sqrt[3]{x}^3 $ ?

b) $f(x) = \sin x$ and $g(x) = x^2$

$$(f\circ g) = f(g(x)) = f(x^2) = \sin (x^2)$$

$$(g\circ f) = g(f(x)) = g(\sin x) = (\sin x)^2 = \sin^2x$$

Am I doing it right? I'm kinda confused whether I have to replace x with the original value or not.

so you meant composition or what does mean $fg$?multiplication? — dato datuashvili, Dec 15 '13 at 15:41

amWhy · Accepted Answer · 2013-12-15T15:50:40.583

5

Yes, you are doing the function compositions correctly. In the first case, since you had doubt on how to write $g\circ f$:

$$g\circ f = g(f(x)) = g(\sqrt[3] x) = \left(\sqrt[3] x\right)^3 = x$$

So, indeed, in the first case, we have $(f\circ g)(x) = (g \circ f)(x) = x$: by definition then, $g(x) = f^{-1}(x)$: $f$ and $g$ are called "inverses" of one another.

Very often, though, as you see in your second pair of functions, $f\circ g \neq g\circ f$.

edited Dec 15 '13 at 15:50

answered Dec 15 '13 at 15:44

amWhy

209,954

Mhm, so you raise the whole function on ^3, right? Not only x. – user95523 Dec 15 '13 at 15:45
Right, exactly. So it turns out that in the first case, we obtain the same result: $f\circ g = g\circ f = x$. – amWhy Dec 15 '13 at 15:46
Ok, sorry for asking too many questions, but, in case we have gf = g(f(x)) and after we replace f(x) with its value, we treat the value inside () of g as "x" and we apply the g(x) to all of it, right? – user95523 Dec 15 '13 at 15:49
right,please look on both answer,@amWhy showed everything clearly,you are welcome – dato datuashvili Dec 15 '13 at 15:49
Yes, you've got precisely the right idea! ;-) – amWhy Dec 15 '13 at 15:49
@amWhy we are happy when we can help someone who want to learn mathematic – dato datuashvili Dec 15 '13 at 15:50
you are welcome,try to use this site regularly,you will study many things – dato datuashvili Dec 15 '13 at 15:53
@amWhy: Could use another TU - you have been way too busy! :-) – Amzoti Dec 16 '13 at 01:52

score 3 · Answer 2 · answered Dec 15 '13 at 16:01

One important remark on notation: $f$ is a function, $f(x)$ is this function applied to $x$, they are not the same thing, no $=$ should ever be between them. The same goes for $f\circ g$. You should not write $f\circ g=f(g(x))$. Where does the $x$ come from? Instead write $$(f\circ g)(x)=f(g(x))=f(x^3)=\sqrt[3]{x^3}=x,\text{ and hence }f\circ g=\mathrm{id}.$$ You can leave off the last part if that is not notation that you have used. Just as well you could write $$(f\circ g)(x)=f(g(x))=\sqrt[3]{g(x)}=\sqrt[3]{x^3}=x.$$ That these yield the same result should give us some confidence in the notation.

Also note that the domains and codomains of the functions should be given. Is $\sqrt[3]{x}$ defined for $x<0$?

Lastly, not everyone agrees with me on this, but I would recommend not to leave off the parentheses in $(f\circ g)(x)$. Indeed, I would read $f\circ g(x)$ as $f\circ(g(x))$ (which does not make sense here). So be on the safe side and write $(f\circ g)(x)$.

score 1 · Answer 3 · answered Dec 15 '13 at 15:45

1

so there is two different case and different result,for example let us suppose that

$f(x)=x^3$

and $g(x)=x^2$

then multiplication is directly $x^3*x^2=x^5$

for composition like $f\circ g=f(g(x))=(x^2)^3$

$g\circ f=g(f(x))=(x^3)^2$

answered Dec 15 '13 at 15:45

dato datuashvili

9,194

Solve the composition $f \circ g=?$ and $g\circ f=?$

3 Answers3