x
pass
x 1
through the point ( 1, 2 ) ? At which points do these tangent lines touch the curve?
PROBLEM 51 (Page 113): How many tangent lines to the curve y 
First, check to see if the point ( 1, 2 ) is on the curve y 
y(1) 
x
:
x 1
x
1
 2 . Thus, the point ( 1, 2 ) is not on the curve y 
.
x 1
2
In general, we have the following picture:
NOTE: The point ( x , f ( x ) ) is the tangent point.
Since the two points of ( x , f ( x ) ) and ( a , b ) are two points on the tangent line,
we can find the slope of the tangent line using algebra:
m tan 
f ( x)  b
x  a
Of course, we can find the slope of the tangent line to the graph of y  f ( x ) at the
point ( x , f ( x ) ) using calculus: m tan  f ( x )
Thus, we have that f ( x ) 
Thus, f ( x ) 
f ( x)  b
x
. For this problem, f ( x ) 
.
x  a
x 1
1( x  1)  x (1)
x 1 x
1
=
=
and
2
2
( x  1)
( x  1) 2
( x  1)
x
x
 2
 2
x  2 ( x  1)
x 1
x 1
x 1
f ( x)  b

=
=
=
=
x  a
( x  1) ( x  1)
x 1
x 1
x 1
x2
x  2x  2
 ( x  2)
=
=
( x  1) ( x  1)
( x  1) ( x  1)
( x  1) ( x  1)
Thus, f ( x ) 
f ( x)  b
 ( x  2)
1



x  a
( x  1) ( x  1)
( x  1) 2
( x  1) ( x  1)   ( x  2 ) ( x  1) 2 
( x  1) ( x  1)  ( x  2 ) ( x  1) 2  0 
( x  1) [ x  1  ( x  2 ) ( x  1) ]  0 
( x  1 ) ( x  1  x 2  3x  2 )  0  ( x  1) ( x 2  4 x  1)  0 
x  1  0 or x 2  4 x  1  0
x  1  0  x   1 . Since the domain of the function f is all real numbers
except  1 , then we do not have a tangent point at  1 .
x  4x  1  0  x 
2
b 
b2  4 a c
2a

4 
16  4 (1) (1)
2

4 
4 
16  4
=
2
4  2 3
12
=
2
2
= 2 
3
x
of the
x 1
tangent lines that pass through the point ( 1, 2 ) . Thus, there are two tangent lines
that pass through the point ( 1, 2 ) .
These are the x-coordinates of the tangent points to the graph of y 
If x   2 
2 
3
3 1

3 , then f (  2 
3  1
2 
3
=
3  1
1 
3
1
3
2
=
3
1
3
2 
3 ,
1 

1
3
1
3
2 
3
3 1
=
3
=
3 1
1
3
=
3 1
3
.
2
Thus, one
3

2  .
3 , then f (  2 
2 
2 
2 3  2  3 


tangent point is   2 

If x   2 
3) 
3) 
2 
2 
2  2 3 
=
3
3  1
=
3  3


. Thus, the other tangent point is   2 

3 ,
1
3
1 
3
1 
=
1 3
2 
2
=
3
=
3

.
2

Maple commands to solve
this problem and draw the
graph.