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

Give an example of a function, g(x), that has a local maximum

at (-3, 3) and a local minimum at (3,-3).

Give an example of a function, g(x), that has a local maximum

at (-3, 3) and a local minimum at (3,-3).

I'm looking for just one function that satisfys both the maximum and minimum

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Since the function cannot be a quadratic (a quadratic has one local extreme), then the function is of 3rd order, at least.

We'll consider a function of 3rd order.

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

We'll impose the constraint that the function has a local maximum at (-3,3).

That means that the x coordinate of the local maximum represents the critical value of function. That means that x = -3 is the root of the first derivative of the function.

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

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

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

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)

Since 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

**The requested function, whose local maximum is at (-3 , 3) and local minimum is at (3 , -3), is: g(x) = x^3/18 - 3x/2.**