Here we look at the bare minimum essentials of the topic needed to solve problems on the GRE.

A function is an expression of the form $f(x)=\text"some expression in x"$, which yields a value given a particular value of x.

e.g. $f(x)=3x+5$, so $f(3)=3*3+5=14$

A function can be defined to take multiple variables. These bind to the appropriate parts of the formula on the RHS.

e.g. $g(x,y)=x+y$, so $g(3,5)=3+5=8$

A**function composition** is designed by applying one function to another, e.g. $f(g(x))$, or to itself e.g. $f(f(x))$

e.g. if $f(x)=3x+2$, and $g(x)=2x+3$, then what are $f(f(2))$, $g(g(2))$, $f(g(2))$ and $g(f(2))$?

The set of values of x over which a function f(x) is defined, is called the**domain** of function f.

The set of values of f(x) that result over the domain of function f, is called its**co-domain**

If $f(-x)=f(x)$ the function f(x) is called an**even function**.

If $f(-x)=-f(x)$ the function f(x) is called an**odd function**.

If a function g(x) can be defined for a given function f(x), such that $g(f(x))=x$, then g(x) is called the**inverse function** of f(x).

In such cases, function f(x) is called an**invertible** function. g(x) is sometimes denoted as $f^{-1}(x)$

e.g. are $f(x)=x^2$ and $g(x)=x^3$ even, or odd functions? What about $h(x)=x^2+x$

if $f(x)=x^2$ then $f(-x)=(-x)^2=(-1)^2*x^2=x^2=f(x)$, so f(x) is an even function.

if $g(x)=x^3$ then $g(-x)=(-x)^3=(-1)^3*x^3=-1*x^3=-1*g(x)$, so g(x) is an odd function.

if $h(x)=x^2+x$, then $h(-x)=(-x)^2+(-x)=x^2-x$. h(-x) is neither equal to h(x) nor to -h(x), so h(x) is neither even nor odd.

A function is an expression of the form $f(x)=\text"some expression in x"$, which yields a value given a particular value of x.

e.g. $f(x)=3x+5$, so $f(3)=3*3+5=14$

A function can be defined to take multiple variables. These bind to the appropriate parts of the formula on the RHS.

e.g. $g(x,y)=x+y$, so $g(3,5)=3+5=8$

A

e.g. if $f(x)=3x+2$, and $g(x)=2x+3$, then what are $f(f(2))$, $g(g(2))$, $f(g(2))$ and $g(f(2))$?

$f(x)=3x+2$, and $g(x)=2x+3$, so:

$f(f(x))=3(3x+2)+2=9x+6+2=9x+8$ and $g(g(x))=2(2x+3)+3=4x+6+3=4x+9$

So $f(f(2))=9(2)+8=26$, and $g(g(2))=4(2)+9=17$

Similarly $g(f(x))=2(3x+2)+3=6x+4+3=6x+7$ and $f(g(x))=3(2x+3)+2=6x+9+2=6x+11$

So again, $g(f(2))=6(2)+7=19$ and $f(g(2))=6(2)+11=23$

The set of values of x over which a function f(x) is defined, is called the

The set of values of f(x) that result over the domain of function f, is called its

If $f(-x)=f(x)$ the function f(x) is called an

If $f(-x)=-f(x)$ the function f(x) is called an

If a function g(x) can be defined for a given function f(x), such that $g(f(x))=x$, then g(x) is called the

In such cases, function f(x) is called an

e.g. are $f(x)=x^2$ and $g(x)=x^3$ even, or odd functions? What about $h(x)=x^2+x$

if $f(x)=x^2$ then $f(-x)=(-x)^2=(-1)^2*x^2=x^2=f(x)$, so f(x) is an even function.

if $g(x)=x^3$ then $g(-x)=(-x)^3=(-1)^3*x^3=-1*x^3=-1*g(x)$, so g(x) is an odd function.

if $h(x)=x^2+x$, then $h(-x)=(-x)^2+(-x)=x^2-x$. h(-x) is neither equal to h(x) nor to -h(x), so h(x) is neither even nor odd.

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