How do you find the critical numbers for #x^4-2x^3-3x^2-5# to determine the maximum and minimum?

Question

William Abdalla · Accepted Answer

Please see the explanation section below,

#f(x) = x^4-2x^3-3x^2_5# is a polynomial, its domain is #(-oo,oo)# #f'(x) = 4x^3-6x^2-6x# is never undefined and is #0# at the solutions to #4x^3-6x^2-6x = 0# Factor out the common factor of #2x# to get #2x(2x^2-3x-3)=0# This leads to two lower degree polynomial equations: #2x=0# #" "# or #" "# #2x^2-3x-3=0# The first has solution #0# and the solutions to the second may be found using the quadratic formula #x= (-b+-sqrt(b^2-4ac))/(2a)#. We get #x=(3+-sqrt33)/4#. #f'(x)=0# at #0#, #" "# #(3+sqrt33)/4#,#" "# and #" "# #(3-sqrt33)/4# All of these solutions are in the domain of #f#, so they are all critical numbers for #f#.

William Abdalla · Answer

undefined To find the critical numbers of $ f(x) = x^4 - 2x^3 - 3x^2 - 5 $, where maximum and minimum points occur, follow these steps:

1. Find the derivative of the function $ f'(x) $.
2. Set $ f'(x) $ equal to zero and solve for $ x $.
3. Determine the values of $ x $ that make $ f'(x) $ undefined.
4. Test each critical number and the endpoints of the interval for maximum and minimum values.

So, first, find $ f'(x) $:
$$ f'(x) = 4x^3 - 6x^2 - 6x $$

Now, set $ f'(x) $ equal to zero and solve for $ x $:
$$ 4x^3 - 6x^2 - 6x = 0 $$
$$ 2x(2x^2 - 3x - 3) = 0 $$

The solutions for $ 2x^2 - 3x - 3 = 0 $ are found using the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

$$ x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-3)}}{2(2)} $$
$$ x = \frac{3 \pm \sqrt{33}}{4} $$

So, the critical numbers are $ x = \frac{3 + \sqrt{33}}{4} $ and $ x = \frac{3 - \sqrt{33}}{4} $.

Next, determine if there are any values of $ x $ that make $ f'(x) $ undefined. Since $ f'(x) $ is a polynomial, it's defined for all $ x $.

Lastly, test each critical number and the endpoints of the interval for maximum and minimum values.