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67 Cards in this Set

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  • Back
definition of a critical value
f'(c)=0 or if f'(c) does not exist than c is called a critical value of f
Test for increasing and Decreasing Functions
1. a function f is increasing on an interval if f'(x) > 0 on that interval
2. A function f is decreasing on an interval if f'(x)<0 on that interval
First Derivative Test= relative minimum
C=critical value of a function f

1.If f'(x) changes from negative to positive at c, then f(c) is a relative minimum of f
First Derivative test= relative maximum
C= critical value of a function f

if f'(c) changes from positive to negative at c, then f (c) is a relative maximum of f
First derivative test= no relative max or min
f'(x) does not change signs at c
relative extrema
relative extrema only occur at critical values but all critical values are not necessarily relative extrema
Extreme Value Theorum
If f is continuous on the closed interval [a,b], then f has botha minimum and a maximum on the interval
Finding extrema on a closed interval
1. find 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 least of these values is the minimum, the greatst is the maximum
Test for concavity=concave up
if f''(x)>0 for all x in an open interval I, then the graph of f is concave upward on I
Test for concavity= concave down
if f''(x)<0 for all x in an open interval I, then the graph of f is concave downward on I
definition of points of inflection
a point P on a curve y=f(x) is called a point of inflection if f is continues on P and the curve changes from concave upward to concave downward or vice versa

If (c, f(c)_ is a point of inflection of the graph of f, then either f''(x)=0 or f is not differentiable at x=c
Second Derivative test= relative minimum
f'(c)=0 and f''(x) exists on an open interval containing c
1. if f''(c)>0, the f(c) is a relative minimum
Second derivative test= relative maximum
f'(c)=0 and f''(x) exists on an open interval containing c
if f''(c) <0, then f(c) is a relative maximum
Second Derivative test=failed
f'(c)=0 and f''(x) exists on an open interval containing c
if f''(c)=0 the test fails use the first derivative test
sketching a curve
1. domain
2. intercepts
3. symmetry
4. asymptote
5. intervals of increasing or decreasing
6. relative maximum and relative minimum
7. points of inflection and concavity
domain
what are the values of x for which f(x) is defined
intercepts
x intercept= let y=0 and solve for x
y intercept= let x=0 and solve for y
symmetry on the y axis
if f(-x)=f(x) then f is an even function and the curve is symmetric about the Y AXIS
symmetry about the origin
if f(-x)=-f(x) then f is an odd function and the curve is symmetric about the origin
symmetry around the x-axis
when replacing y by -y yields an equivalent equation the curve is symetric about the x axis
maximum or minimum value of a certain quantity is called...
optimization
optimization guidlines
1. draw a picture
2. assign a variable to the unknown quantity and write an equation for the quantity that is to be maximized or minimized
3. Use the given information to find a relationship between the variables. use these equations to eliminate all but one variable in the equation
4. use the first and second derivative tests to find critical points
tangent line approximation
line tangent to a curve at a point is the line that best approximates the curve near that point
tangent line at point (c, f(c)) as approximation to the curve y=f(x) when x is near c
tangent line approximation equation
y-f(c)=f'(c)*(x-c) or y=f(c)+f'(c)*(x-c)
derivative of √u
1/(2√(u))du/dx
chain rule
derivative of 1/u
(-1/u^2)du/dx
chain rule
derivative of lnu
1/u du/dx
chain rule
derivative of logₐu
(1/lna)*(1/u)*du/dx
chain rule
derivative of e^u
e^u*du/dx
chain rule
the derivative of 7^u
(ln7)*7^u*du/dx
chain rule
the derivative of sin(u)
cos(u)
the derivative of cos(u)
-sin(u)
the derivative of tan(u)
sec²u
the derivative of cot(u)
-csc²u
the derivative of sec(u)
sec(u)tan(u)
the derivative of csc(u)
-csc(u)cot(u)
the derivative of arcsin(u)
1/√(1-u²)
the derivative of arccos(u)
-1/(√(1-u²))
the derivative of arctan(u)
1/1+u²
the derivative of arccot(u)
-1/(1+u²)
the derivative of arcsec(u)
1/(u*√(u²-1))
the derivative of arccsc(u)
-1/(u*√(u²-1))
∫u^n
1/(n+1)*u^(n+1)+C
∫1/u
ln∣u∣+C
∫7^u
(1/ln7)*a^u+C
∫tan(u)
ln∣sec(u)∣+C
∫cot(u)
ln∣sin(u)∣+C
∫sec(u)
ln∣sec(u)+tan(u)∣+C
∫csc(u)
ln∣csc(u)-cot(u)∣+C
express f'(x) as a limit
f'(x)= lim as h-->0 f(x+h)-f(x)/h
express f'(a) as a limit
f'(x)= lim as h-->0 f(a+h)-f(a)/h
express f'(a) as limit approaches a
f'(a)= lim as x--> a f(x)-f(a)/(x-a) because x=a+hand h=x-a
when is a function differentiable
a function f is differentiable at a only if f'(a) exisits. If a function is diferentiable at x=a, the f is continues at x=a, just because it is continuous does not mean it is differentiable
instantaneous rate off change
f'(x)
when is f(x) not differentiable
1. when the graph has a corner
when the graph is not continuous
when the graph has a vertical tangent
ln(xy)

simplify
lnx+lny
lnx^p
simplify
plnx
lnx/y

simplify
lnx-lny
e^lnx
simplify
x
slope of tangent line
remember m=...?
derivative
tangent line is horizontal w/ respect to dy/dx
dy/dx=0
tangent line is vertical w/ respect to dy/dx
when denominator in expression for dy/dx is 0
f^(-1)'(x)
1/(f'(f^-1(x))
e^1
e
ln(e)
1
-sin(arcsinx)
-x
Guidlines forsolving related rate problems
1. read problem carfully--draw diagram if possible.
2. Name the variables and constants
3. Write an equatioon that relates thevariables whose rates of change are give. Write anequation whoserates ofchange are to be determined.
4. combine two or more equations to get a single variable
5. usechainrule to differentiate bothsides with respect to t
6.substittute the given numerical information into the resulting equation and solve for the unknownrate