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Differentiation help please..urgent.?

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for the function y= -2x^3 - 3x^2 + 4x + 2, the values of the y coordinate and the gradient at x=1 are...

Given that f(x)=2x^3 - 3x^2 -12x, the value(s) of x for which f ' (x)=0 is/are...

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  1. 1)

    y = -2x³ - 3x² + 4x + 2

    Differentiate this (which will give you the slope of 1)

    dy/dx = -6x² - 6x + 4

    -6x² - 6x + 4 = 1

    -6x² - 6x + 3 = 0

    Solve the quadratic using the quadratic formula:

    x = [-b ± √(b² - 4ac)] / 2a

    x = [6 ± √((-6)² - 4*-6*3)] / 2*-6

    x = [6 ± √(36 + 72)] / -12

    x = (6 ± √108) / -12

    x = (6 ± 6√3) / -12

    x = (1 ± √3) / -2

    Plug this x-value into your equation given:

    y = -2x³ - 3x² + 4x + 2

    And get your 2 y values.

    =================================

    f(x) = 2x³ - 3x² -12x

    f'(x) = 6x² - 6x - 12 = 0

    6x² - 6x - 12 = 0

    Solve the quadratic:

    x = [-b ± √(b² - 4ac)] / 2a

    x = [6 ± √(6² - 4*6*-12)] / 2*6

    x = [6 ± √(36 + 288)] / 12

    x = (6 ± √324) / 12

    x = (6 ± 18) / 12

    x = 24/12

    x = -12/12

    x = 2

    x = -1

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