How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #y = x^3 - 3x^2 - 9x +15#?

Answer 1

At #x=3#, we have a relative minima and at #x=-1# we have a relative maxima.

A function is increasing, when the value of its derivative at that point is positive and is decreasing, when the value of its derivative at that point is negative.

At maxima and minima, the value of derivative is #0#, as at maxima, the slope of the curve stops increasing and starts declining and at minima the slope of the curve stops declining and starts increasing.

Further, at maxima, second derivative is negative and at minima, second derivative is positive.

Derivative of #f(x)=x^3-3x^2-9x+15# is #f'(x)=3x^2-6x-9#, which is zero when #3x^2-6x-9=0# or #x^2-2x-3=0#. Further second derivative #f''(x)=6x-6#
Factorizing #x^2-2x-3=0# becomes #(x-3)(x+1)=0# and hence minima or maxima is at #x=3# or #x=-1#. As at #x=3#, #f''(x)=12# we have a minima and at #x=-1# #f''(x)=-12# we have a maxima.

However, as elsewhere the functions takes lower value than the minima or higher value than the maxima these are relative minima and maxima.

graph{x^3-3x^2-9x+15 [-10, 10, -25, 25]}

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Answer 2

To determine where the function ( y = x^3 - 3x^2 - 9x + 15 ) is increasing or decreasing, and to find the relative maxima and minima:

  1. Find the first derivative of the function.

  2. Set the first derivative equal to zero and solve for ( x ) to find critical points.

  3. Test the intervals between critical points using the first derivative test to determine where the function is increasing or decreasing.

  4. Find the second derivative of the function.

  5. Use the second derivative test at the critical points to determine where the function has relative maxima or minima.

  6. First derivative: [ y' = 3x^2 - 6x - 9 ]

  7. Set ( y' ) equal to zero and solve for ( x ): [ 3x^2 - 6x - 9 = 0 ] [ x^2 - 2x - 3 = 0 ] [ (x - 3)(x + 1) = 0 ] [ x = 3 \quad \text{or} \quad x = -1 ]

  8. Test intervals:

    • Test the interval ( (-\infty, -1) ): Pick ( x = -2 ), ( y'(-2) = 3(-2)^2 - 6(-2) - 9 = 3(4) + 12 - 9 = 15 > 0 ), so the function is increasing in this interval.
    • Test the interval ( (-1, 3) ): Pick ( x = 0 ), ( y'(0) = 3(0)^2 - 6(0) - 9 = -9 < 0 ), so the function is decreasing in this interval.
    • Test the interval ( (3, \infty) ): Pick ( x = 4 ), ( y'(4) = 3(4)^2 - 6(4) - 9 = 15 > 0 ), so the function is increasing in this interval.
  9. Second derivative: [ y'' = 6x - 6 ]

  10. Use the second derivative test:

    • At ( x = -1 ): ( y''(-1) = 6(-1) - 6 = -12 < 0 ), so there is a relative maximum at ( x = -1 ).
    • At ( x = 3 ): ( y''(3) = 6(3) - 6 = 12 > 0 ), so there is a relative minimum at ( x = 3 ).

Therefore, the function is increasing on ( (-\infty, -1) \cup (3, \infty) ), and decreasing on ( (-1, 3) ). The relative maximum occurs at ( x = -1 ) and the relative minimum occurs at ( x = 3 ).

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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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