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Calculus 1
Worksheet #13 Even
Review for Test #3
2
f ( x)  3x  4  f '( x)  lim
h 0
4
f(x)=
2
x4
f '( x)  lim
2
2

 x  h  4 x  4
h 0
 lim
h 0
6
8
10
3( x  h)  4  (3x  4)
3x  3h  4  3x  4
3h
 lim
 lim
 lim 3  3
h

0
h

0
h
h
h h 0
 lim
 x  h  4
2
( x  4)
2



 x  h   4 ( x  4) x  4  x  h   4
h 0
h
2( x  4)  2   x  h   4 
= lim 
 x  h  4  ( x  4)
h 0
h
h

2 x  8  2 x  2h  8
2h
2
2
 lim
 lim

h

0
h

0
h  x  h  4  ( x  4)
h  x  h  4  ( x  4)
 x  h  4  ( x  4) ( x  4)2
1
1
d
d
1
3  2 x   3  2 x  2   3  2 x  2   2  
dx
dx
2
2
1
y  3x 3  4 x 2  2 
1
3  2x
1
1
1
1
dy 2
1
  3x 3   4 x 2  2 x 3  2 x 2
dx 3
2
1
 2 x 
1

1  x2
x2
1  x 2 (1)  x 
1  x 2 (1)  x  1  x 2  2  2 x  


2
1
x
dy
1  x2 
2

 2 1 x  
y


2
2
dx
1  x2
1  x2
1  x2
1  x2




1  x2  x2
1  x2
1  x2
12
14
16

1
1  x 
2
1  x2
or
1
1  x 
2 3
1
 y '   x 2  y ''  2 x 3  y '''  6 x 4  y iv  24 x 5  y iv (1)  24(1) 5  24
x
y  5 y'  0
y
x
x
x
2x
3x
 1 
f ( x)  x x  f '( x)  x 





  x 1 
1
2 x
2 x 2 x
2 x
2 x 
3
3x
x
3x x 3 x
3 1 3



or smarter! f ( x)  x x  x 2  f '( x)  x 2 
x
2x
2
2
2
2 x
x
18
20
22
y   x 2  2  x 1  2  
d 2
x  2  x 1  2    x 1  2   2 x    x 2  2   x 2   2  4 x  1  2 x 2  1  2 x 2  4 x

dx
2 5
d
2 5
2 10
y  3x 2   2  (3x 2   2 )  6 x  2  3
x x
dx
x x
x
x
3( x  1) 2  2 
6( x  1) 2
 x 1 
 x  1   ( x  1)(1)  ( x  1)(1) 
f ( x)  
  f '( x) 

  f '( x) 
  f '( x)  3 
 
( x  1) 2
( x  1) 2  ( x  1) 2 
( x  1) 4
 x 1
 x 1 

24 Find the x and y intercepts of the line that is tangent to the curve y = x3 at the point (-2,-8)
3
2
Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions
Calculus 1
Worksheet #13 Even
Review for Test #3
x  int@(4 / 3, 0)
y  x3  y '  3x 2  y '(2)  12  y  8  12( x  2)  y  12 x  16 
y  int@(0,16)
26
1
Find the slope of the normal to f(x) =2x3+x2-1 at the point x= .
2
5
2
1
1
1 3
f '( x)  6 x 2  2 x  f '    6    2     1 
Slope of the normal 
2
5
2
4
2 2
28
1
Graph the function whose equation is g(x)=
then find and graph the tangent at point (3, f(3))
x2
1
1
g ( x) 
 g '( x) 
 g '(3)  1  y  1  1( x  3)  y   x  4
x2
( x  2)2
30 Graph the function y  2 x 2  4 x  1 then find and graph the tangent line at the point where x=1.
y  2 x 2  4 x  1; tangent at x=1
y  2( x 2  2 x  1)  1  2  y  2( x  1) 2  3  y '  4 x  4  y '(1)  8  y  5  8( x  1)  y  8 x  3
32 Find the points on the curve x  1 where the tangent is parallel to the y-axis.
1
y  x  1  y ' 
 undefined @(1, 0)
2 x 1
34
dy
Find
for y   x  1  x 2  5 x  1 then find the slope of the tangent when x=3
dx
dy
dy
y  ( x  1)( x 2  5 x  1) 
 ( x  1)(2 x  5)  ( x 2  5 x  1)(1) 
 2 x2  5x  2 x  5  x2  5x  1
dx
dx
dy
dy
 3x 2  8 x  6  (3)  27  24  6  45
dx
dx
36 Find the slope of the tangent line and the normal line to the curve y  x x  1 at the point where x=1.
dy
dy
x
dy
1
2 4 2 5 2
 1 
 x
 x  1(1) 

 x  1  (1) 
 2 



dx
dx 2 x  1
dx
4
4
4
2 2
 2 x 1 
2 2
5 2
The slope of the normal 
5
4
38 If the slope of a tangent line is 5 then what is the slope of the normal line to the same curve at the same point?
1
5
2
40
 x 1 
Find the equation of the normal line to the curve f ( x)  
 at the point where x=2.
 x 1
The slope of the tangent 
2
 x 1 
 x 1
 x  1   ( x  1)(1)  ( x  1)(1) 
 x  1   2 
f ( x)  
  2
  f ( x)  
  f '( x)  2 


2
2 
( x  1)
 x 1
 x 1 
 x 1  
 x  1   ( x  1) 


2
4( x  1)
4(3)
1
1
 f '(2) 
 12  Slope of normal   y  9  ( x  2)
3
( x  1)
1
12
12