Expert: Paul Klarreich - 10/22/2010

Question
Why does the function x^4+3x^3-2 have a relative min at (-2.25, -10.54) with no absolute minimum but the function x^4+5x^3+3x^2-4x has an absolute minimum at (-3.18, -15.47) and a relative minimum at (0.25, -3.25).  It seems that they are similar graphs with the same interval so they should either both have a relative min and absolute min or they should both have a relative min with no absolute min.  What am I not seeing?  

Answer
Questioner:   Shelly
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  relative extrema
Question:  Why does the function x^4+3x^3-2 have a relative min
at (-2.25, -10.54) with no absolute minimum

>> I don't think that is right.  It has an absolute (and
relative) min at that point, but no other rel. min.

but the function x^4+5x^3+3x^2-4x has an absolute minimum at (-
3.18, -15.47) and a relative minimum at (0.25, -3.25).

>> yes, that is so.  
It seems that they are similar graphs with the same interval so
they should either both have a relative min and absolute min or
they should both have a relative min with no absolute min.  What
am I not seeing?
...............................
They don't have to have all similar properties.
They DO have in common:
A. Both quadric, so so there will be an absolute min OR absolute

max.
B. Both have a0 > 0, so there will be an absolute min, but no

absolute max.

The first:

f(x) = x^4 + 3x^3 - 2

has critical points as follows:

f'(x) = 4x^3 + 9x^2 = 0

x^2(4x - 9) = 0

x = 0, a double root
x = 2.25, as you found.

At x = 0, you also have:

f''(x) = 12x^2 + 18x

You will find inflection points at x = - 3/2,

AND AT x = 0

When a critical point is also an inflection point, it may or may
not be a rel min (or max).

.........................
Now the second:

g(x) = x^4+5x^3+3x^2-4x

g'(x) = 4x^3 + 15x^2 + 6x - 4

I don't think it has its rel min at 0.25, but somewhere nearby.  
Anyway, all three c.p.'s are different.