Download Section 2.1

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Section 2.1: The Derivative and the Tangent Line Problem
Practice HW from Larson Textbook (not to hand in)
p. 87 # 1, 5-23 odd, 59-62
Tangent Lines
Recall that a tangent line to a circle is a line that touches the circle only once.
If we magnify the circle around the point P
we see that slope of the slope of the tangent line very closely resembles the slope of the
circle at P. For functions, we can define a similar interpretation of the tangent line slope.
Definition: The tangent line to a function at a point P is the line that best describes the
slope of the graph of the function at a point P. We define the slope of the tangent line to
be equal to the slope of the curve at the point P.
Examples of Tangent Lines:
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Consider the following graph:
Slope of Secant line
between the points
=
(x, f(x)) and (x+h, f(x+h))
As h→0, the slope of the secant lines approach that of the tangent line of f at x = a.
Slope of
tangent line = m =
of f at (x, f(x))
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Definition: Given a function y  f (x) , the derivative of f, denoted by f  , is the function
defined by
f ( x)  lim
h0
f ( x  h)  f ( x )
,
h
provided the limit exists.
Facts
1. For a function y  f (x) at a point x = a, f (a ) gives the slope of the tangent line to
the graph of f at the point (a, f (a)) .
2. f (a ) represents the instantaneous rate of change of f at x = a, for example,
instantaneous velocity.
Example 1: Use the definition of the derivative to find the derivative of the function
f ( x)  3x  2 .
Solution:
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Example 2: Use the definition of the derivative to find the derivative of the function
f ( x)  2 x 2  x  1.
Solution:
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Example 3: Find the equation of the tangent line to the curve f ( x)  x 3  3x 2  4 at the
point (1, 2).
Solution:
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Example 4: Use the definition of the derivative to find f (x ) if f ( x)  1  x .
Solution: Using the limit definition of the derivative, we see that
f ( x)  lim
h 0
 lim
h 0
f ( x  h)  f ( x )
h
1 x  h  1 x
h


f ( x)  1  x
 Note that


f ( x  h)  1  x  h 

( 1 x  h  1 x) ( 1 x  h  1 x)
h 0
h
( 1 x  h  1 x)
 lim
 Multiply numerator and denominato

 by the conjugate of the numerator
(1  x  h)  1  x  h 1  x  1  x  h 1  x  (1  x)
h 0
h ( 1 x  h  1 x)
 lim
 FOIL the numerator and 


 multiply denominato r 
1 x  h 1 x
h 0 h ( 1  x  h  1  x )
(Simplify and distribute - sign to 1  x)
h
h 0 h ( 1  x  h  1  x )
(Simplify)
 lim
 lim
 lim
h 0
1
1 x  h  1 x
(Simplify by canceling h terms)

1
1 x  0  1 x
(Substitut e h  0 to evaluate the limit)

1
1 x  1 x
(Simplify)

1
2 1 x
r


(Simplify by combining like terms)
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Example 5: Find the derivative using the definition and use the result to find the equation of the
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line tangent to the graph of f ( x) 
at the point (0, -1).
x 1
Solution: To find the equation of any line, including a tangent line, we need to know the line’s
slope and a point on the line. Since we already have a point on line, we must find the tangent line’s
slope, which is found using the derivative. Using the limit definition of the derivative, we see that
f ( x)  lim
h 0
f ( x  h)  f ( x )
h
1
1

x

h

1
x
1
 lim
h 0
h
1
1


, f ( x  h) 
 Note that f ( x) 

x

1
x

h

1


1
[ x  1]
1 [ x  h  1]

( x  h  1) [ x  1] ( x  1) [ x  h  1]
 lim
h 0
h
The common demonimato r is ( x  h  1)( x  1) 
( x  1)
( x  h  1)

( x  h  1)( x  1) ( x  1)( x  h  1)
 lim
h 0
h
Rewrite
( x  1)  ( x  h  1)
( x  h  1)( x  1)
 lim
h 0
h
( x  1  x  h  1)
( x  h  1)( x  1)
 lim
h 0
h
fractions in terms of common denominato r 
(Combine fractions over common denominato r )
(Distribut e - sign)
h
( x  h  1)( x  1)
 lim
h 0
h
h
( x  h  1)( x  1)
 lim
h
h 0
1
h
1
 lim

h  0 ( x  h  1)( x  1) h
( Simplify)
(Write h as
h
)
1
(Take reciprocal of denominato r and multiply)
1
(Cancel h terms)
h  0 ( x  h  1)( x  1)
1
1
1



( x  0  1)( x  1) ( x  0  1)( x  1)
( x  1) 2
 lim
(Simplify to get the derivative )
Continued on next page
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Using f ( x)  
1
( x  1)2
, we can now find the slope at the give point (0, -1).
Slope at the point
1
1
1
(0,1)
 m  f (0)  

   1 .
2
2
1
(0  1)
(1)
( x  0)
Using m  1, we see that from the slope intercept equation y  mx  b
that
y  x  b
To find b, use the fact that at the point (0, -1), x  0 and y  1 . Thus
 1  0  b
giving b  1. Thus, the equation of the tangent line is:
y  x 1.
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Differentiability and Points where the Derivative Does Not Exist
Note: The derivative of a function may not exist a point.
Fact: If a function is not continuous at a point, its derivative does not exist at that point.
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is not continuous at x = 1. This implies that f (1) will not
x 1
1
exist. Note that f ( x)  
, which computationally says f (1) does not exist.
( x  1) 2
For, example, f ( x) 
Note! However, a function may still be continuous at a point but the derivative may still
not exist.
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Example 10: Use the definition of the derivative to demonstrate that f (0) does not exist
for the function f ( x)  x .
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Fact: In general, a function that displays any of these characteristics at a point is not
differentiable at that point
Cases
1. The function is not continuous at a point – it has a jump, break, or hole in the graph at
that point.
2. The function has a sharp point or corner at a point.
3. The function has a vertical tangent at a point.
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Example 11: Determine the point)s on the following graph where the derivative does not
exist. Give a short reason for your answer.
Solution:
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