Answer :

To find the extreme values of the function [tex]y = 12x^5 - 45x^4 + 40x^3[/tex], we need to find its critical points where the first derivative is zero or undefined and then determine the nature of these critical points by using the second derivative test.

Step 1: Find the first derivative

First, let's find the derivative of the function with respect to [tex]x[/tex]:

[tex]y' = \frac{d}{dx}(12x^5 - 45x^4 + 40x^3)[/tex]

[tex]y' = 60x^4 - 180x^3 + 120x^2[/tex]

Step 2: Find the critical points

To find the critical points, set the derivative equal to zero and solve for [tex]x[/tex]:

[tex]60x^4 - 180x^3 + 120x^2 = 0[/tex]

Factor out the greatest common factor:

[tex]60x^2(x^2 - 3x + 2) = 0[/tex]

Set each factor to zero:

  1. [tex]60x^2 = 0[/tex] leads to [tex]x = 0[/tex].

  2. [tex]x^2 - 3x + 2 = 0[/tex] is a quadratic equation which can be factored as:

    [tex](x - 1)(x - 2) = 0[/tex]

    This gives [tex]x = 1[/tex] and [tex]x = 2[/tex].

The critical points are [tex]x = 0, 1, 2[/tex].

Step 3: Use the second derivative test

Calculate the second derivative:

[tex]y'' = \frac{d^2}{dx^2}(12x^5 - 45x^4 + 40x^3)[/tex]

[tex]y'' = 240x^3 - 540x^2 + 240x[/tex]

To determine the nature of each critical point, substitute the value of [tex]x[/tex] into the second derivative:

  • For [tex]x = 0[/tex]:

    [tex]y''(0) = 240(0)^3 - 540(0)^2 + 240(0) = 0[/tex]

    The second derivative test is inconclusive. Further analysis such as the first derivative test is needed.

  • For [tex]x = 1[/tex]:

    [tex]y''(1) = 240(1)^3 - 540(1)^2 + 240(1)[/tex]

    [tex]y''(1) = 240 - 540 + 240 = -60[/tex]

    Since [tex]y''(1) < 0[/tex], [tex]x = 1[/tex] is a local maximum.

  • For [tex]x = 2[/tex]:

    [tex]y''(2) = 240(2)^3 - 540(2)^2 + 240(2)[/tex]

    [tex]y''(2) = 1920 - 2160 + 480 = 240[/tex]

    Since [tex]y''(2) > 0[/tex], [tex]x = 2[/tex] is a local minimum.

The extreme values occur at [tex]x = 1[/tex] (local maximum) and [tex]x = 2[/tex] (local minimum). To find the actual extreme values, substitute [tex]x = 1[/tex] and [tex]x = 2[/tex] back into the original function [tex]y[/tex].

  • [tex]y(1) = 12(1)^5 - 45(1)^4 + 40(1)^3 = 12 - 45 + 40 = 7[/tex] (local maximum)

  • [tex]y(2) = 12(2)^5 - 45(2)^4 + 40(2)^3 = 384 - 720 + 320 = -16[/tex] (local minimum)

Thus, the local maximum value is [tex]y = 7[/tex] at [tex]x = 1[/tex], and the local minimum value is [tex]y = -16[/tex] at [tex]x = 2[/tex].

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