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What are the extreme points of f(x) = 3x^3 - 16x^2 + 2x

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lxsptter | Student, Undergraduate | (Level 2) Valedictorian

Posted April 7, 2013 at 4:37 PM via web

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What are the extreme points of f(x) = 3x^3 - 16x^2 + 2x

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 7, 2013 at 4:49 PM (Answer #1)

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For a function f(x), at the extreme points the value of f'(x) = 0. If the solution of f'(x) = 0 is a, and the value of f''(a) is negative a point of maximum lies at x = a; if the value of f''(a) is positive a point of minimum lies at x = a.

For f(x) = 3x^3 - 16x^2 + 2x, f'(x) = 9x^2 - 32x + 2

The solution of the equation 9x^2 - 32x + 2 = 0 is `x1 = (32 + sqrt(32^2 - 36))/18`   = `16/9 + sqrt 247/9` and `x2 = (32 + sqrt(32^2 - 36))/18` = 1`6/9 - sqrt 247/9`

f''(x) = 18x - 32

`f''(16/9 + sqrt 247/9) = 2*sqrt 247` which is positive

`f''(16/9 - sqrt 247/9) = -2*sqrt 247` which is negative.

For the function f(x) = 3x^3 - 16x^2 + 2x, the maximum value lies at x = `16/9 - sqrt 247/9` and the minimum value lies at x = `16/9 + sqrt 247/9`

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