Answer :

To find the extreme values of the function [tex]y = 12x^5 - 45x^4 + 40x^3[/tex], we need to follow these steps:


  1. Find the derivative of the function

    The first step is to take the derivative of the function with respect to [tex]x[/tex]. The derivative, [tex]y'[/tex], will help us find the critical points.

    [



y' = \frac{d}{dx}(12x^5 - 45x^4 + 40x^3) = 60x^4 - 180x^3 + 120x^2
]


  1. Set the derivative equal to zero and solve for [tex]x[/tex]

    To find the critical points, set the derivative equal to zero:

    [



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

Factor out the greatest common factor, which is [tex]60x^2[/tex]:

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

This equation results in two parts: [tex]60x^2 = 0[/tex] and [tex]x^2 - 3x + 2 = 0[/tex].

From [tex]60x^2 = 0[/tex], we get:

[tex]x = 0[/tex]

Solve the quadratic equation [tex]x^2 - 3x + 2 = 0[/tex]. Factor the quadratic as:

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

Therefore, we get:

[tex]x = 1 \, \text{and} \, x = 2[/tex]


  1. Evaluate the original function at these critical points

    Now that we have the critical points [tex]x = 0, 1, \text{and} \, 2[/tex], evaluate [tex]y[/tex] at each of these points to find the extreme values.

    [tex]x = 0:[/tex]
    [



y = 12(0)^5 - 45(0)^4 + 40(0)^3 = 0
]

[tex]x = 1:[/tex]
[tex]y = 12(1)^5 - 45(1)^4 + 40(1)^3 = 12 - 45 + 40 = 7[/tex]

[tex]x = 2:[/tex]
[tex]y = 12(2)^5 - 45(2)^4 + 40(2)^3 = 12 \times 32 - 45 \times 16 + 40 \times 8 = 384 - 720 + 320 = -16[/tex]


  1. Determine the extreme values

    The extreme values of the function are at the points [tex](0, 0)[/tex], [tex](1, 7)[/tex], and [tex](2, -16)[/tex]. The point [tex](1, 7)[/tex] is a local maximum, and the point [tex](2, -16)[/tex] is a local minimum.



In summary, the function has a local maximum at [tex](1, 7)[/tex] and a local minimum at [tex](2, -16)[/tex]. These are the extreme values of the function.

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