RAC functions Flashcards
function
let A and B be sets. A function from A to B is a rule that associates every element of A to exactly one element of B.
set of inputs
domain
set of outputs
codomain
image/range
for every element from A, xЄA then the element of B associated to x by f is called the image
difference between codomain and image/range
codomain are the possible outputs
range/image are the actual outputs
image definition (function f: A->B)
the subset of B whose elements are the images of elements A. denoted by F(A)
f: A->B g: C->D f=g if…
- A=C
- B=D
- f(x) = g(x) for all xЄA
graph of symbol
Γ
image letters
f(A) = {f(x): xЄA}
graph letters
Γf = {(x,y)ЄAxB: y=f(x)}
how to tell if a graph is a graph of a function
if one input has multiple outputs it is not a function (draw a vertical line, if it crosses two points its not the graph of a fucntion)
constructing function via restriction definintion
Let f:A->B. Let S⊆A. The restriction of f to S is the function f|s: S->B defined by f|s(x) = f(x) for all values of xЄS
what is restriction
taking a smaller domain to only show part of the function
constructing a function via composition definition
Let f:A->B and g:C->. Assume that f(A)⊆C (if B=C then definitely true). Then the composition of g and f is the function g∘f: A-> D defined by g∘f(x) = g(f(x)) for all values of xЄA
indentity function definition
Let A be a set. The identity function of A is the function idₐ: A->A defined by idₐ(x) = x for all values of xЄA
what happens if you do composition with an identity function
nothing
Constant Function definition
A function f:A->B is said to be constant if there exists bЄB such that f(x)=b for all xЄA
what do functions have to be to be invertible
one-to-one
invertibility definition
Let f: A-> B. We say that f is invertible if there exists a function g:B->A such that for all xЄA and yЄB f(x) = y <=> g(y) =x
Injective (one-to-one)
if for all x,x’ЄA: x=x’ => f(x)!=f(x’) (all points have distinct images)
if every image point has at most one preimage point in A.
surjective
f(A) = B
if every image point has at least one preimage in A
bijective
both surjective and injective.
if every image point has exactly one preimage in A
g: B->A is the inverse of f:A->B iff
g∘f = idₐ and f∘g = idb
real valued function of a real variable
f: A->ℝ , where A⊆ℝ