Semester 1 Vocab Flashcards
dsin(x) =
cos(x)
dcos(x) =
-sin(x)
dtan(x) =
sec^2(x)
dcsc(x) =
-csc(x)cot(x)
dsec(x) =
sec(x)tan(x)
dcot(x) =
-csc(x)
dsin^-1(x) =
1/(1-x^2)^1/2
dcos^-1(x) =
-1/(1-x^2)^1/2
dtan^-1(x) =
1/(1+x^2)
if f is continuous on a closed interval [a, b], the f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers c and d in [a, b].
Extreme Value Theorem
to find the absolute maximum and minimum values of a continuous function f 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 endpoints of the interval
3. the largest of the values from steps 1 & 2 is the absolute maximum
Closed Interval Method
if f is differentiable on the interval [a, b], then there exist a number c between a and b such that:
f’(c) = (f(b) - f(a))/ (b - a)
Mean Value Theorem
suppose c is a critical number of a continuous function f.
a) if f changes from + to - at c, then c is a local maximum
b) if f changes from - to + at c, then c is a local minimum
c) if f does not have a sign change at c, then there is not a local max or min at c
1st Derivative Test
suppose c is continuous near c.
a) if f’(c) = 0 and f’‘(c) > 0, f has a local min at c
b) if f’(c) = 0 and f’‘(c) < 0, f has a local max at c
2nd Derivative Test
suppose function f and function b are differentiable and g’(x) can’t equal 0 near a (except possibly at a)
L Hop