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

### 1 Answer | Add Yours

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`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes