How do you find the exact relative maximum and minimum of the polynomial function of #g(x) = x^3 - 3x^2 - 9x +1#?

Question

Emilia Aguilar · Accepted Answer

#P_("max") ->(x,y) ->(-1,6)#
#P_("min")->(x,y)->(3,-26)#

#(dy)/(dx)=3x^2-6x-9 = 0#

Divide throughout by 3 giving

#x^2-2x-3=0#

#(x+1)(x-3)=0#

#x=-1" or "+3#
'~~~~~~~~~~~~~~~~~~~~~~
#(d^2y)/(dx^2)=6x-6#

At #x=-1 ; (d^2y)/(dx^2)<0 => "maximum"#

At #x=+3 ; (d^2y)/(dx^2)>0 => "minimum"#

By substitution you find

#P_("max") ->(x,y) ->(-1,6)#
#P_("min")->(x,y)->(3,-26)#

Charles Anctil · Answer

undefined To find the exact relative maximum and minimum of the polynomial function $ g(x) = x^3 - 3x^2 - 9x + 1 $, follow these steps:

1. Find the critical points by taking the derivative of the function and setting it equal to zero.
2. Determine the nature of each critical point using the second derivative test.
3. Verify the endpoints of the interval if it exists.
4. Identify the relative maximum and minimum points.

Here are the detailed steps:

1. Find the derivative of $ g(x) $:
$$ g'(x) = 3x^2 - 6x - 9 $$

2. Set $ g'(x) $ equal to zero and solve for $ x $ to find the critical points:
$$ 3x^2 - 6x - 9 = 0 $$
$$ x^2 - 2x - 3 = 0 $$
$$ (x - 3)(x + 1) = 0 $$
$$ x = 3, -1 $$

3. Find the second derivative of $ g(x) $:
$$ g''(x) = 6x - 6 $$

4. Test the critical points using the second derivative test:
   - At $ x = 3 $:
     $$ g''(3) = 6(3) - 6 = 12 > 0 $$
     The second derivative is positive, so $ g(x) $ has a relative minimum at $ x = 3 $.
   - At $ x = -1 $:
     $$ g''(-1) = 6(-1) - 6 = -12 < 0 $$
     The second derivative is negative, so $ g(x) $ has a relative maximum at $ x = -1 $.

5. Verify the endpoints of the interval (if applicable):
   Since there are no specified intervals, you don't need to verify endpoints.

Therefore, the exact relative maximum is at $ (x, g(x)) = (-1, 7) $ and the exact relative minimum is at $ (x, g(x)) = (3, -17) $.