01 Graphs Flashcards
(fg)-1
(fg)-1 = f-1g-1
f f-1 (x)
f f-1 (x)=x
Domain D
the range given in the qn
Usually x axis
Range
Extent of Y axis
() vs []
() not inclusive
[] including that value
how to get f-1 from f(x)
make x the subject
replace x with f-1
Test for Existence of Functions
only if any vertical line x = k, where k is a constant, k â Df
cuts the graph at one and only one point.
Test for inverse functions
to check if it is one-to-one
use the horizontal line test
f is one-one if every horizontal line y = k, k â R
cuts the graph at one point.
Relationship between a Functions and its Inverse
The point(s) of intersection of g and g<sup>-1 </sup>lie on the line y = x Function and its Inverse are reflected images of each other in the line y = x
Inverse functions range/domain
R<sub>f-1</sub> = D<sub>f</sub> R<sub>f</sub> = D<sub>f-1</sub>
Test for composite function
R1 must be a subset to D2
Domain of composite function gf
Dgf = Df
Range of composite function gf
take Df domain of first
insert into domain of second Dg
Find Rg range of second
Yea thats the range of composite Rgf woo
f f-1 (x) = f-1 f(x) = x
Inverse composite domain
The domain of f-1 f(x) is Df
The domain of ff-1 (x) is Df -1
Domain follows the first function
Finding asymptotes
Oblique: Long division if improper frac
Vertical: Let denom be 0
divide coeff of numerator and denominator
Eqn of circles
(đĽ â â) 2+(đŚ â đ) 2= đ 2 , where đ â 0
Eqn of ellipses
(đĽ â â)2/đ2 + (đŚ â đ)2/đ2 = 1
Eqn of hyperbola and asymptotes
(đĽ â â)2/đ2 - (đŚ â đ)2/đ2 = 1
đŚ = Âą đ/đ (x-h) + k
Scaling
replace by y/2 to x2
replace by 2y to halve
Reflection
-y = f(x) flips x axis y = f(-x) flips y axis
Translation
y-1 shift by 1 unit positive y direction
y+1 shift 1 unit neg y direction
y = F(|x|)
Keep all right hand side +x and copy over
drawing f(x) to fâ(x)
horizontal asymptotes become y=0
vertical asymptotes remain
f(x) to 1/f(x) drawing
asymptotes become x intercepts
x intercepts become asymptotes
max pt to min pt
min pt to max pt
oblique symptotes to horizonal y=0