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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
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If f(x1)>f(x2) and x1< x2, then the function f(x) is __________
the graph is________ |
decreasing, downhill
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If f(x1)>=(x2) for every x1, x2, then the function f(x) is _________
the graph is________ |
constant, horizontal
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F(c) is a maximum value of f if...
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f(x)<_ f(c) for all x on the interval I
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F(c) is a minimum value of f if...
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f(x)>_ f(c) for all x on the interval I
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F(c) is a local maximum of f if there is an open interval containing c such that:
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f(x)<_ f(c) for all x on the interval (a,b)
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F(c) is a local minimum of f if there is an open interval containing c such that:
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f(x)>_f(c) for all x on the interval (a,b)
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steps for finding critical numbers
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1) differentiate the function
2) find the c values for which f'(c)= 0 or f'(c) dne |
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Steps for finding the local extrema of f on (a,b)
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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 |
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Rolle's theorem
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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 |
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Strategy for using Rolle's theorem
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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) |
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Mean Value Theorem
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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) |
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Strategy for using Mean Value theorem
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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) |
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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 |
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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 |
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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 |
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Strategy for 1st derivative test
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1)prove continuity and differentiability
2) differentiate 3)find CN 4)determine intervals 5)determine signs 6)determine local extremum |
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Concavity:
graph is concave up if... |
f''(c)>0
tan line is below graph |
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Concavity:
graph is concave down if... |
f''(c)<0
tan line is above graph |
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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 |
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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 |
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Second Derivative test
the test is inconclusive if... |
f is differentiable on an open interval containing c and
f''(c)0 |
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Strategy for second derivative test
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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 |
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definition: inflection point
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point at which the function changes from concave up to concave down
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how to find inflection points
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set the second derivative equal to 0 and solve
check with a sign chart |
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Strategy for an application problem
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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 |
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Strategy for an infinity limit if degrees are the same
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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 |
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lim as x-> infinity of c/ (x^k)=
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0
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lim as x-> negative infinity of c/ (x^k)=
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0
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Strategy for an infinity limit if degrees are different
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Divide function and rewrite as a product
cancel values that go to zero |