Download Math 1205 Test 3 Review: Logarithms, Derivatives, Rates, Graphing and more Study notes Calculus in PDF only on Docsity!
REVIEW SHEET Math 1205 Test 3 Sect. 3.8 - 4.5; Wrkshts 7&
- Logarithms: logarithms are exponents.
ln x = c means e
c
= x ln x defined for x > 0
Properties:
- ln( AB ) = ln A + ln B 4. ln e
x
= x
- ln
A
B
= ln A − ln B 5. e
ln x
= x
- ln A
p
= p ln A 6. ln1 = 0 and ln e = 1
- Last of the Derivative Rules:
d
dx
(ln x ) =
x
d
dx
log
a
x
x ln a
d
dx
(ln u ) =
u
du
dx
d
dx
log
a
u
u ln a
du
dx
Problems:
f ( x ) = ln(2 − x ) 2.
f ( x ) = x ln x 3.
f ( t ) =
1 + ln t
1 − ln t
F ( x ) = e
x
ln x 5.
y = x
x
y = (sin x )
x
- Related Rates: Scheme:
- draw picture; label constants and variables
- indicate, Given:, Find:
- set up an equation relating variables and constants for all time
- find D
t
- substitute and solve
Problems: p 260, #7 and hwk problems.
- Linearization:
- L ( x ) = f ( a ) + f ′( a ) *( x − a ) is the equation of the tanline to f ( x ) at a
called the standard approximating fctn; linearization of f ( x ) at a
- Differential:
dx is an independent variable
dy = f ′( x ) ⋅ dx
Uses: 1. approximate the curve f ( x )
- estimate change in area or volume
- evaluate an error in measurement.
Problems:
- Find the linearization L ( x ) of the function at a :
f ( x ) = x
3
, a = 1
- Find the linearization L ( x ) of the function at a :
f ( x ) = x
3
, a = − 8
- (a) Find the differential
dy and (b) evaluate
dy for the given values of
x and dx for
y = x
2
- (a) Find the differential
dy and (b) evaluate
dy for the given values of
x and dx for
y = 1 − x , x = 0 , dx = 0.
- pg 268 #
- pg 268 #
- Mean Value Theorem:
Given f(x) continuous on [a,b] and differentiable on (a,b),
then for at least one c in (a,b)
f ( b ) − f ( a )
b − a
= f
/
( c )
Problems:
- p 295 #
- p 295 #
- Extrema of Functions:
Maximum and Minimum:
A function, continuous on a closed interval, always has an abs max and abs min.
They occur at Critical Points (where
f ( c ) = 0 or
f ( c ) dne ) or at endpoints.
Problems:
- Find the absolute maximum and absolute minimum values of f on the given interval:
f ( x ) = 3 x
2
− 12 x + 5 , 0 , 3 [ ]
- Find the absolute maximum and absolute minimum values of f on the given interval:
f ( x ) = x
2
x
Local Maxima and Minima:
Occur at Critical Pts where the following is true:
using 1st derivative test:
where f ′( x ) changes from + to - f ( x ) has a local max
where f ′( x ) changes from - to + f ( x ) has a local min
or 2nd derivative test
if f ′′( x ) > 0 f(c) is a local min
if f ′′( x ) < 0 f(c) is a local max
- Concavity:
Given f ( x ) continuous and differentiable.
- f ( x ) increasing when f ′( x ) > 0
decreasing when f ′( x ) < 0
- f ( x ) concave up when f ′′( x ) > 0 ( f ′( x ) is increasing)
concave down when f ′′( x ) < 0 ( f ′( x ) is decreasing)
- f ( x ) has pt of inflection where fctn changes concavity.
- (Review) Limits and Asymptotes:
Vertical Asymptote: x = a is a V.A. of f ( x ) if lim
x → a
±
f ( x ) = ±∞
a point of discontinuity is therefore a potential V.A.
Horizontal Asymptote: y = b is a H.A. of f ( x ) if lim
x →±∞
f ( x ) = b
- Scheme for Sophisticated Graphing:
- f ( x ) find intercepts, asymptotes, y values
- f ′( x ) find critical pts, increasing and decreasing intervals, local max and min
- f ′′( x ) find possible pts of inflection, intervals of concavity, pts of inflection
Problems:
- pg 304 #
