A.  Graph the function ƒ(x) = x³ - 4x + 2.



B.  Find the domain and range of ƒ, showing all work or explaining your rationale.

ƒ(x) is a polynomial function of odd degree.

All polynomial functions are defined for all values of x and have domain
(,)

All odd degree polynomial functions have both domain and range
(,).


C.  Find the derivative of ƒ, showing all work.
 
ƒ(x) = x³ - 4x + 2
ƒ'(x) = 3x² - 4 

D.  Find the slope of the graph of ƒ at x = 0, showing all work.
 
ƒ(x) = x³ - 4x + 2
ƒ'(x) = 3x² - 4 
ƒ'(0) = 3(0)² - 4
ƒ'(0) = -4

The slope of the graph at x = 0, which is the slope of the 
green line below which is tangent to the curve at (0,2)



E.  Let ƒ represent the position of an object with respect to time that is moving along a line. Identify when the object is moving in the positive and negative directions and when the object is at rest, showing all work.

ƒ(x) = x³ - 4x + 2

f(x) = 3x² - 4f(x) = (3x² - 4)}

The object is moving in the positive direction when 3x² - 4 > 0. That is,

Find critical numbers of 3x² - 4 

3x² - 4 = 0
    3x² = 4
     x² = 
      x = 
      x =   approximately ±1.1545

These critical numbers are when the object is at rest, because
the derivative of f with respect to time is 0 there.



Choose test value -2 which is less that 
Substitute in 3x² - 4
3x² - 4
3(-2)² - 4
3(4) - 4
12 - 4
8

That's positive, so the object is moving in the positive direction
when x < 

Choose test value 0 which is between the two critical numbers
 and 

Substitute in 3x² - 4
3x² - 4
3(0)² - 4
-4

That's negative, so the object is moving in the negative direction
when  < x < 

Choose test value 22 which is greater that 
Substitute in 3x² - 4
3x² - 4
3(2)² - 4
3(4) - 4
12 - 4
8
That's positive, so the object is moving in the positive direction
when x > 

Answer:  

The object is moving in the positive direction when 
x <  and when x > 

The object is moving in the negative direction when 
 <  x < 

The object is at rest when x =  and when x = 

Edwin

f(x)=

Here is the graph:





 does not exist because

 =  = 

Therefore lim, f(x),
"x->1","" )}}} does not exist because the denominator becomes 0 but the
numerator does not become 0, but becomes -1. Therefore, there is a
non-removable infinite discontinuity at x=1. That's where ther is a
vertical asymptote. 

 =  =  


There is no point at (2,1). However since both numerator and
denominator approach 0, the limit might exist and there may be a
removable discontinutity there indicated by the small circle on
the graph.

 =  =  =  =  =  =  = 1

So the limit exists at 2 and equals 1. However f(2) is not defined.

So there is a removable discontinuity at x=2 

Sometimes this is called "a hole in the graph".

So there is a non-removable infinite discontinuity at x=1,
and a removable discontinuity at x=2`


Edwin