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I was doing home work but I am stuck with some questions, here is the question with subparts,

Let $A=B=C=\Bbb R$, and let $f:A\to B,g:B\to C$ be defined by $f(a)=a-1$ and $g(b)=b^2$. Find

$\text{(a)}\,\,\,\,(f\circ g)(2)\qquad\qquad\text{(b)}\,\,\,\,(g\circ f)(2)$

$\text{(c)}\,\,\,\,(g\circ f)(x)\qquad\qquad\text{(d)}\,\,\,\,(f\circ g)(x)$

$\text{(e)}\,\,\,\,(f\circ f)(y)\qquad\qquad\,\text{(f)}\,\,\,\,(g\circ g)(y)$

Now I am stuck with what does $(2)$, $(x)$ and $(y)$ mean in these questions? For example, (a) $(f \circ g)(2)$ What is $(2)$ here?

Please help me out.


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These are parameters to the functions. Recall that a function is an object that takes input (in your case, some number) and produces output. For example, $g$ takes a number and returns its square. –  Yuval Filmus Aug 26 '11 at 15:00
Function composition ($\circ$) is an operation that takes two functions and produces a function obtained by running the function in succession. For example, composing $f$ and $g$ in one order produces a function $h(w) = w^2 - 1$, and in the other order it produces the function $i(w) = (w-1)^2$. –  Yuval Filmus Aug 26 '11 at 15:02
You can also compose a function with itself, for example $g \circ g$ means apply $g$ and then apply it again, i.e. square the input and then square it again. So in total $(g \circ g)(T) = g(T^2) = (T^2)^2 = T^4$. –  Yuval Filmus Aug 26 '11 at 15:03
Thanks alot. That was full of knowledge :) –  Shen Xu Aug 26 '11 at 15:11

1 Answer 1

up vote 6 down vote accepted

2 in $(f \circ g)(2)$ is where the function $g$ is evaluated. $$(f \circ g)(2)=f(g(2))=f(2^2)=f(4)=4-1=3.$$

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Thankyou so very much for the answer. –  Shen Xu Aug 26 '11 at 15:22
@fahad: Glad I could help. –  Américo Tavares Aug 26 '11 at 15:32

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