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30 Cards in this Set

  • Front
  • Back
If f(x1)<f(x2) and x1<x2, then the function f(x) is __________

the graph is________
increasing, uphill
If f(x1)>f(x2) and x1< x2, then the function f(x) is __________

the graph is________
decreasing, downhill
If f(x1)>=(x2) for every x1, x2, then the function f(x) is _________

the graph is________
constant, horizontal
F(c) is a maximum value of f if...
f(x)<_ f(c) for all x on the interval I
F(c) is a minimum value of f if...
f(x)>_ f(c) for all x on the interval I
F(c) is a local maximum of f if there is an open interval containing c such that:
f(x)<_ f(c) for all x on the interval (a,b)
F(c) is a local minimum of f if there is an open interval containing c such that:
f(x)>_f(c) for all x on the interval (a,b)
steps for finding critical numbers
1) differentiate the function
2) find the c values for which f'(c)= 0 or f'(c) dne
Steps for finding the local extrema of f on (a,b)
1) find the critical numbers
2)calculate f(c) for each critical number
3)calculate f(a) and f(b)
4) the absolute max and min are the largest values calculated in steps 2 and 3
Rolle's theorem
IF
-f is continuous on [a,b]
-f is differentiable on (a,b)
-f(a)= f(b)
THEN
there must be a critical number on the interval (a,b) such that f'(c)=0
Strategy for using Rolle's theorem
1) show that the function satisfies the conditions of continuity, differentiability, and equality of f(a) and f(b)
2)differentiate the function
3)set f'= 0 and solve for x
4) use the x value above and solve for f(x)
Mean Value Theorem
IF
-f is continuous on [a,b]
-f is differentiable on (a,b)
THEN
there exists a number c such that:
f(b)-f(a)= f'(c)(b-a)
Strategy for using Mean Value theorem
1) show that the function satisfies the conditions of continuity and differentiability
2) calculate f(a), f(b), and f'(c)
3) plug values into f(b)-f(a)= f'(c)(b-a)
4) solve for c
5) find f(c)
First Derivative test

f(c) is a local max if:
Let f be continuous at c and differentiable on an open interval containing c and let e>0.

f'(c-e)>0 and f'(c+e)<0
aka: f' changes from positive to negtive
First Derivative test

f(c) is a local min if:
Let f be continuous at c and differentiable on an open interval containing c and let e>0.

f'(c-e)<0 and f'(c+e)>0
aka: f' changes from negative to positive
1st derivative test

f(c) is not a local extremum
Let f be continuous at c and differentiable on an open interval containing c and let e>0.

f'(x)>0 or f'(x)<0 for every x except x=c
Strategy for 1st derivative test
1)prove continuity and differentiability
2) differentiate
3)find CN
4)determine intervals
5)determine signs
6)determine local extremum
Concavity:
graph is concave up if...
f''(c)>0
tan line is below graph
Concavity:
graph is concave down if...
f''(c)<0
tan line is above graph
Second Derivative test
x has a local max at c if...
f is differentiable on an open interval containing c and F(C)=0!!
f''(c)<0
Second Derivative test
x has a local min at c if...
f is differentiable on an open interval containing c and F(c)=0!!
f''(c)>0
Second Derivative test
the test is inconclusive if...
f is differentiable on an open interval containing c and
f''(c)0
Strategy for second derivative test
1) differentiate
2) find CN
3)find f''(x)
4) find f''(c)
5)decide concave up or down
6) determine local max and min
definition: inflection point
point at which the function changes from concave up to concave down
how to find inflection points
set the second derivative equal to 0 and solve
check with a sign chart
Strategy for an application problem
1)read
2)draw picture
3)write down facts and relationships
4) determine which variable is to be maximized or minimized and write it as a function of one other variable
5)find critical values of function found in 4 and test to see if they are max or min
6) check to see if extremes occur at the endpoints of function obtained in four
Strategy for an infinity limit if degrees are the same
multiply top and bottom by a form of one, a form which has x to the highest degree in the denominator.
Cancel values that go to zero
lim as x-> infinity of c/ (x^k)=
0
lim as x-> negative infinity of c/ (x^k)=
0
Strategy for an infinity limit if degrees are different
Divide function and rewrite as a product
cancel values that go to zero