What is the extreme value of 2x^3+3x^2-12x+5?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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The extreme values of a function occur at the points where the derivative is equal to 0.

f(x) = 2x^3 + 3x^2 - 12x + 5

=> f'(x) = 6x^2 + 6x - 12

6x^2 + 6x - 12 = 0

=> x^2 + x - 2 = 0

=> x^2 + 2x - x - 2 = 0

=> x(x + 2) - 1(x + 2) = 0

=> (x - 1)(x + 2) = 0

x = 1 and x = -2

f(x) = -2 for x = 1 and f(x) = 25 for x = -2.

The extreme values are x1 = -2 and x2 = 25.

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To determine the extreme values of the given function, we'll have to calculate the critical points of the expression.

For this reason, we'll determine the first derivative, since the critical points are the roots of the first derivative.

f'(x) = (2x^3+3x^2-12x+5)'

f'(x) = 6x^2 + 6x - 12

We'll put f'(x) = 0:

6x^2 + 6x - 12 = 0

We'll divide by 6:

x^2 + x - 2 = 0

We'll apply the quadratic formula:

x1 = [-1+sqrt(1 + 8)]/2

x1 = (-1+sqrt9)/2

x1 = (-1+3)/2

x1 = 1

x2 = (-1-3)/2

x2 = -2

The extreme values of the function are:

f(1) = 2*1^3+3*1^2-12*1+5

f(1) = 2 + 3 - 12 + 5

f(1) = -2

f(-2) = 2(-2)^3+3(-2)^2-12(-2)+5

f(-2) = -16+ 12+24 + 5

f(-2) = 25

The extreme values of the function are: f(1) = -2 and f(-2) = 25.

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