# Radical_ Irrational_ Too Complex to be Real_ by yurtgc548

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```									  Tangents

Calc AB- Section2.7
If we want to find the tangent line at a point P of a curve, we
consider a nearby point and find the SLOPE of the secant line
_______
through P and Q
Qx0  h, f x0  h
y                                   y

Px0 , f x0                                    QQ
Q
Q
P
P
h
x
x
x0                      Tangent line at P
as h goes to zero
x0  h
Definitions: TANGENT LINE
Px0 , f x0 
The slope of the curve y  f  x  at the point ____________ is
the number

f x0  h   f x0 
m    lim
h 0
(Provided the limit exists)
h
Line Through
The tangent line to the curve at P is the ____________ P with
the slope m defined above
Find the line tangent to the curve y  f  x  at the point  x0 , y0 
3
Example 1: Find the line tangent to the hyperbola f  x 
at the point  3,1
x

f x0     f x0  h 
Step 1: Calculate ______ and _________

f 3   1           f 3  h  
3                            3
3                           3 h
Step 2: Find the _________ of the tangent line
SLOPE
3
1        3  3  h           h               1
3 h     lim                lim             lim
h 0 h3  h    h 0 3  h 
lim
h 0    h       h 0 h3  h 

1        1
EQUATION                                   
3  0 3
Step 3: Find the ____________
of the tangent line.
1
y 1     x  3
3
Rates of Change

AVERAGE RATE OF CHANGE
The average rate of change is the SLOPE of the SECANT line.
______       ________
It’s an _________________!
APPROXIMATION
Through the points
y f x0  h   f x0          x0 , f x0  x0  h, f x0  h

x          h

INSTANTANEOUS RATE OF CHANGE
The instantaneous rate of change is the ________ of the
SLOPE
TANGENT
____________ line.
It’s __________!
EXACT
f x0  h   f x0 
lim
h 0          h
Example 2:
The distance traveled, in feet, by a free falling rock from rest
is given by the function f  t   16t 2 , where t  seconds.

a) Estimate the rocks speed in the interval from t  1 to t  1.1

d f 1  h   f 1 161  h   161 16h 2  2h   16h  2
2       2

                                   
t         h                    h              h
Since t goes from 1 to 1.1 the change in time is 0.1 = h

160.1  2  33.6 ft sec

The estimated speed of the rock at t =1 is 33.6 ft/sec
b) Find the actual speed of the rock at t  1

161  h   16        161  2h  h 2   16        16  32h  16h 2  16
2
lim                   lim                          lim
h 0        h          h 0          h               h 0          h

h32  16h 
 lim                lim 32  16h   32  16  0  32 ft sec
h 0      h         h 0

The ACTUAL speed of the rock at t =1 is 32 ft/sec
Example 3: Find the slope of the tangent line at x  0
for the function f  x   x 1 3
f 0  h   f 0        h1 3  0         1
lim                     lim           lim 2 3  
h 0         h            h 0    h       h 0 h
Vertical Tangents
A curve y  f  x  has a vertical tangent at the point x  x0
f x0  h   f x0 
if lim
h 0
    or  
h
Example 4: Find the slope of the tangent line at x  0
for the function g  x   x
23

f 0  h   f 0          h2 3  0         1
lim                     lim              lim 1 3  
h 0         h            h 0             h 0 h
h

1
lim 1 3  
h 0 h

No vertical tangent at x = 0 because the limit DNE

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