1. Jul 25, 2012

### lionel messi.

Hi all, can someone please give description about the use of composite and inverse functions with examples...? any help will be appreciated :!!)

2. Jul 25, 2012

### HallsofIvy

"Composition of functions" refers to appling to functions in succession. For example, if $f(x)= x^3$ (whatever x is, cube it) and $g(x)= x+ 2$ (whatever x is, add 2) then composition g(f(x)) means- what ever x is, first cube it, then add 2. Here, $g(f(x))= x^3+ 2$. Composition of functions is NOT "commutative"- that is, changing the order may give a different result: g(f(x)), here, means "first add 2 then cube the result: $f(g(x))= (x+ 3)^3= x^3+ 9x^2+ 27x+ 27$.

Of course, whenever we do an operation, we would like to be able to "undo" it. That is exactly what the "inverse" function does: If we write $f^{-1}(x)$ to mean its inverse function, $f^{-1}(f(x))= x$ and $f(f^{-1}(x))= x$, With $f(x)= x^3$, $f^{-1}(x)= \sqrt[3](x)= x^{1/3}$. That is, the "inverse" of cubing is the cube root. $f(f^{-1}(x)= (\sqrt[3](x))^3= x$ or $f(f^{-1}(x))= (x^{1/3})^3= x$ and $f^{-1}(f(x))= \sqrt[3](x^3)$ or $f^{-1}(f(x))= (x^3)^{1/3}= x$.

Similarly if g(x)= x+ 3, then $g^{-1}(x)= x- 3$. That is, the inverse of "add three" is "subtract 3". $g(g^{-1}(x))= (x- 3)+ 3= x$ and $g^{-1}(g(x))= (x+ 3)- 3= x$.

More generally, we can think of "y= f(x)" as a set of pairs of numbers, f= {(x,y)}= {(x, f(x))}. And then the inverse just swaps "x" and "y"- the pairs corresponding to (y, x): $f^{-1}= (y, x)= (y, f^{-1}(y))$

You need to be aware that many functions do NOT have inverses! For example, the function $f(x)= x^2$ does not have an inverse because both f(-2) and f(2) are equal to 4 so there is no single number we can call "$f^{-1}(4)$. What we often do in that case is restrict the domain so the function is "one to one"- but then we have a different function. If we define $f(x)= x^2$ with x restricted to be non-negative, then we can say that $f^{-1}(x)= \sqrt{x}$. And if we define $g(x)= x^2$ with x restricted to be non-positive, then $g^{-1}(x)= -\sqrt{x}$.

3. Jul 25, 2012

### lionel messi.

Thanks a bunch!