Give an example of a function, g(x), that has a local maximum at (−3, 3) and a local minimum at (3, −3).

The function g(x) is of 3rd order or larger, since a
quadratic function cannot have more than one extreme, either a minimum or a maximum
point.

g(x) = ax^3 + bx^2 + cx +
d

Since the function has a local maximum at (-3,3), then x
= -3 is canceling the first derivative of the
function.

We'll differentiate with respect to
x.

g'(x) = 3ax^2 + 2bx +
c

g'(-3) = 27a - 6b + c

But
g'(-3)=0 => 27a - 6b + c = 0 (1)

Now, we'll
calculate g(-3)=3.

g(-3) = -27a + 9b - 3c + d
<=> -27a + 9b - 3c + d = 3 (2)

g'(3) = 0
<=> 27a + 6b + c = 0 (3)

g(3) = -3
<=> 27a + 9b + 3c + d = -3 (4)

We'll equate
(1)=(3):

27a - 6b + c = 27a + 6b +
c

We'll eliminate like
terms:

-12b = 0 => b =
0

We'll add (2) + (4):

-27a +
9b - 3c + d + 27a + 9b + 3c + d = 3 - 3

We'll substitute b
by 0 and we'll eliminate like terms:

2d =
0

d = 0

We'll substitute b and
d by 0 in 27a + 9b + 3c + d = -3

27a + 3c =
-3

We'll divide by 3:

9a + c =
-1 (5)

We know that 27a + 6b + c = 0 and b = 0
=>

=> 27a = -c => 9a = -c/3
(6)

We'll replace 9a by
(6):

-c/3 + c = -1

-c + 3c =
-3 => 2c = -3 => c = -3/2

9a = 3/2*3
=> 9a = 1/2 => a = 1/18

A
possible function g(x), whose local maximum is at (-3 , 3) and local minimum is at (3 ,
-3), is the 3rd order function: g(x) = x^3/18 -
3x/2.