unit 1 test Flashcards
HA rules
degree of numerator < degree of denominator: y = 0
degree of numerator = degree of denominator divide leading coefficient of numerator/ denominator
degree of numerator > degree of denominator: no HA
how to find limits (regular) algebraically
plug in c into the equation
if you get 0/0 (indeterminate form), factor equation
then replug in c and that’s the answer
find limit graphically
find domain(where the denominator equals 0)
holes: when roots in denominator and numerator cross out
VA:set denominator equal to 0(doesn’t include hole)
points of discontinuity: any roots in denominator
3 types of behavior where limit DNE
jump discontinuity: f(x) approached different # from right side of x=c than the left side
f(x) increases or decreases w/o bond (+/- infinity)
oscillation
lim 3
x->2
3
rationalizing technique
multiply by conjugate
conjugate (sq rt. x) + 4 = (sq rt. x) - 4
squeeze theorem
how does graph look
3 lines sandwiched together
solving for point that connects all 3 lines in the middle
squeeze theorem
the actual formula/theorem
h(x) = f(x) = g(x) for all x in an open interval containing c; except possibly c itself:
lim h(x) = L = lim g(x) x-> c x-> c
then lim f(x) exists and is also equal to L
x-> c
squeeze theorem
solve algebraically
find lim cos x
x-> infinity
find y valué limits of graph
ex. -1/x = cos x = 1/x
then find the limit of first half then second half lim h(x) lim g(x) x-> c x-> c ex. lim -1/x = 0 (1st half) x-> infinity
find limit then of f(x) if h(x) equals g(x)
then identify and state “by Squeeze Theorem”
finding limits of infinity
figure out HA of f(x)
the HA is the answer
if no HA, the limit DNE
find continuity at a point
function is continuous when these 3 conditions are met:
f(c) is defined
lim f(x) exists x->c
lim f(c) = f(c) x-> c
ALL 3 conditions HAVE to be TRUE
types of discontinuities
hole: removable discontinuity
the one with a hole then a defined point at the same x- value is also considered a hole
jump: jump discontinuity
what are the functions that are continuous at any point IN THEIR DOMAIN
polynomial, rational, radical, basic, and trig functions(sin,cos, tan)
approaching from the left
-
approaching from the right to
+
continuity on a closed interval
definition/ theorem
a function is continuous on the closed interval [a,b] when f is continuous on the open interval (a,b)
and
lim f(x) = f(a) x->a+
lim f(x) = f(b) x->b-
continuity of a composite function
if g is continuous at c and f is continuous at g(c) then the composite function given by (f of g)(x) = f(g(x)) is continuous at c
Intermediate Value Theorem
definition
if f is continuous on the closed interval [a,b], f(a) is not equal to f(b) and k is any number between f(a) and f(b), then there is at least one number c in [a,b] such that f(c) = k
summary: used to locate zeroes of function that are continuous in the closed interval
intermediate value theorem
how to solve w these problems
[0,1]
plug in a(first number in closed bracket) into og equation
f(0)= -1
then plug in b(2nd number in closed bracket) into og equation
f(1) = 2
if a is less than 0 and his greater than 0
state “ since f(0) <0 and f(1) > 0, by the intermediate value theorem, there exists at least a number c [0,1] such that f(c) = 0
for what value of k is the function f(x) insert piece wise function continuous at x = c
make it so both pereciese parts equal each other
plug in 6 and solve for k