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MATA30

Final Review

QuestionAnswer
lim [f(x) + g(x)] x->a lim f(x) + lim g(x) x->a x->a
lim [f(x) - g(x)] x->a lim f(x) - lim g(x) x->a x->a
lim [cf(x)] x->a c lim f(x) x->a
lim [f(x)*g(x)] x->a lim f(x) * lim g(x) x->a x->a
lim f(x) x->a ---- g(x) lim x->a f(x) ---- lim g(x) x->a
lim n x->a [f(x)] _ _ n | lim f(x)| where n is +integer | x->a |
lim c x->a c
lim x x->a a
lim n x->a x n a where n is a positive integer
lim n-rt(x) x->a n-rt(a) where n is + integer if n is even, we assume that a>0
lim n-rt[(f(x)] x->a n-rt(lim f(x)) x->a if n is even, we assume that lim x->a f(x) > 0
lim f(x) = L x->a if and only if lim f(x) = lim f(x) x->a- x->a+
L'HOSPITAL's RULE f and g are dif'ble on open intrvl I containing a(excpt maybe at a) AND g'(x)<>0. Suppose that lim x-a f(x)=0 & lim x-a g(x)=0 or lim x-a f(x)= +- inf & lim x-a g(x)= +- inf Then lim x->a f(x)/g(x) = lim x-a f'(x)/g'(x) if limit on the R exist(or =
Even function (Cosine) f(-x) = f(x)
Odd function (Sine) f(-x) = -f(x)
Intermediate Value Theorem 1. f(x) is continuous on [a,b] 2. N is any number between f(a) and f(b) , where f(a) <> f(b) Result: There exists a number c in (a,b) such that f(c) = N
Removable discontinuity a discontinuity that could be removed by redefining the function at a single value.
Infinite discontinuity a discontinuity a at which the right or left hand limit is infinity or negative infinity
Continuous from the right a function f(x) is continuous from the right at a number a if lim x->a+ f(x) = f(a)
Continuous from the left a function f(x) is continuous from the left at a number a if lim x->a- f(x) = f(a)
jump discontinuity a discontinuity at which the value of the function jumps. lim x->a+ f(x) = c + lim x->a- f(x) for some non-zero constant c
continuous at an endpoint of an interval a function is continuous at an endpoint of an interval if it is defined on only one side of an endpoint and the function is continuous from that side
continuous on an interval a function is continuous on an interval if it is continuous at every point in the interval
Closed Interval Method To find the abs max and min values of a continuous function on a closed interval [a,b] 1. Find the values of f at the critical numbers of f in (a,b) 2. Find the values of f at the interval endpoints 3. largest value is abs max, smallest value is abs mi
optimization problem a problem, often in differential calculus, in which a function is maximized or minimized
local minimum a function f has a local minimum at c if f(c) <= f(x) for all x in some open interval containing c
local maximum value a number f(c), where c is in the domain of a function f(x), such that f(c) >= f(x) for all x in some open interval containing c.
critical number a number c in the domain of f such that either f'(c) = 0 or f'(c) does not exist
local maximum a function f has a local maximum at c if f(c) >= f(x) for all x in some open interval containing c
minimum value a number f(c), where c is in the domain of a function f(x), such that f(c) <= f(x) for all x in the domain of f
maximum value a number f(c), where c is in the domain of a function f(x), such that f(c) >= f(x) for all x in the domain of f
absolute / global maximum a value c in the domain, D, of the function f such that f(c) >= f(x) for all x in D; the x-coordinate of the highest point on the graph of f(x), though there may be many points with this height.
extreme values the maximum and minimum values of a function f
Extreme Value Theorem 1. f is continuous on a closed interval [a,b] Result: f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers c and d in [a,b]
Increasing / Decreasing test a) if f'(x) > 0 on an interval, then f is increasing on that interval b) if f'(x) < 0 on an interval, then f is decreasing on that interval
Concave downward 1. If the graph of f lies below all of its tangents on an interval Then f is concave downward on I
Second Derivative test 1. f'' is continuous near c. a) if f'(c) = 0 and f''(c) > 0, then f has a local minimum at c. b) if f'(c) = 0 and f''(c) < 0 then f has a local minimum at c
Inflection point A point P on a curve y=f(x) at which f is continuous and at which f changes concavity
Concave upward 1. If the graph of f lies above all of its tangents on an interval Then f is concave upward on I
First derivative test 1. c is a critical number of a continuous function f a) if f' changes from + to - at c, then f has a local max at c b) if f' changes from - to - at c, then f has a local min at c c) if f' does not change sign at c, then f has no max or min at c
Concavity test a) if f''(x) > 0 for all x in an interval I, then the graph of f is concave upward on I. b) if f''(x) < 0 for all x in an interval I, then the graph of f is concave downward on I
Rolle's Theorem Conditions: 1. f is a continuous function on the closed interval [a,b] 2. f is differentiable on (a,b) 3. f(a) = f(b) Result: There is a number c in (a,b) such that f'(c) = 0
Mean Value Theorem Conditions: 1. f is a continuous function on the closed interval [a,b] 2. f is differentiable on (a,b) Result: There is a number c in (a,b) such that f'(c) = [f(b) - f(a)] / b-a
Created by: xjoaniex