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

Please see the explanation section below,

This leads to two lower degree polynomial equations:

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

- Find the derivative of the function ( f'(x) ).
- Set ( f'(x) ) equal to zero and solve for ( x ).
- Determine the values of ( x ) that make ( f'(x) ) undefined.
- 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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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