How do you find the intervals of increasing and decreasing using the first derivative given #y=x(x^2-3)#?

Answer 1

Increasing: #(-oo, -1), (1, oo)#
Decreasing: #(-1, 1)#

Expand to avoid having to use the product rule.

#y= x^3 - 3x#

Use the power rule to differentiate.

#y' = 3x^2 - 3#
Find the critical values. These will occur when the derivative equals #0# or is undefined. There are no undefined points on a quadratic function.
#0 = 3x^2 - 3#
#0 = 3(x^2 -1)#
#0 = x^2 - 1#
#0 = (x + 1)(x- 1)#
#x = -1 and 1#
We have three separate intervals we have to check here. There is #(-oo, -1), (-1, 1) and (1, oo)#.

Pick test points. If when substituted into the derivative the derivative is negative, the function is deceasing over that interval. If the derivative is positive, the function is increasing over that interval.

Test Point 1: #x = -2#
#y' = 3x^2 - 3 -> y' = 3(-2)^2 - 3 = 3(4) - 3 = 9#
#:.#The function is increasing over #(-oo, -1)#.
Test Point 2: #x = 0#
#y' = 3x^2 - 3 -> y' = 3(0)^2 - 3 = 3(0) - 3 = -3#
#:.#The function is decreasing over #(-1, 1)#
Test Point 3: #x = 2#
#y' = 3x^2 - 3 -> y' = 3(2)^2 - 3 = 3(4) - 3 = 9#
#:.# The function is increasing over #(1, oo)#.

Hopefully this helps!

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Answer 2

To find the intervals of increasing and decreasing using the first derivative, follow these steps:

  1. Find the first derivative of the function ( y = x(x^2 - 3) ).
  2. Set the first derivative equal to zero and solve for ( x ) to find critical points.
  3. Test the intervals between the critical points by choosing test points and evaluating the sign of the first derivative.
  4. If the first derivative is positive in an interval, the function is increasing in that interval. If it's negative, the function is decreasing.

Let's proceed with these steps:

  1. Find the first derivative: [ y' = \frac{d}{dx} (x(x^2 - 3)) ] [ y' = (x^2 - 3) + x(2x) ] [ y' = x^2 - 3 + 2x^2 ] [ y' = 3x^2 - 3 ]

  2. Set the first derivative equal to zero and solve for ( x ): [ 3x^2 - 3 = 0 ] [ x^2 = 1 ] [ x = \pm 1 ]

So, the critical points are ( x = -1 ) and ( x = 1 ).

  1. Test intervals:

    • Test ( x = 0 ) (interval ( (-\infty, -1) )): [ y' = 3(0)^2 - 3 = -3 ] (negative)
    • Test ( x = \frac{1}{2} ) (interval ( (-1, 1) )): [ y' = 3\left(\frac{1}{2}\right)^2 - 3 = -\frac{3}{4} ] (negative)
    • Test ( x = 2 ) (interval ( (1, \infty) )): [ y' = 3(2)^2 - 3 = 9 ] (positive)
  2. Interpretation:

    • The function is decreasing on intervals ( (-\infty, -1) ) and ( (-1, 1) ).
    • The function is increasing on interval ( (1, \infty) ).
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Answer from HIX Tutor

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.

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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