Question

How do you find the relative extrema for `f(x)=x^3-4x^2+x+6`?

Answer 1

relative min at `( 4/3 +(sqrt(13))/3, 70/27 - (26 sqrt(13))/27)`

relative max at `(4/3 - (sqrt(13))/3, 70/27 + (26 sqrt(13))/27)`

Explanation:

Given: `f(x) = x^3 - 4x^2 + x + 6`

To find relative extrema first find the first derivative :

`f'(x) = 3x^2 - 8x + 1`

Find the critical value(s) by setting `f'(x) = 0`

`f'(x) = 3x^2 - 8x + 1 = 0`

Use the quadratic formula to find the critical value(s):

`x = (8 +- sqrt(8^2 - 4(3)(1)))/(2*3) = 4/3 +- sqrt(52)/6 = 4/3 +-( sqrt(13))/3`

Find critical points :

`f(4/3 +(sqrt(13))/3) = 70/27 - (26 sqrt(13))/27 ~~-.8794`

`f(4/3 -(sqrt(13))/3) = 70/27 + (26 sqrt(13))/27 ~~6.0646`

Use 2nd derivative test since it is easy to find the 2nd derivative of this function:

`f''(x) = 6x - 8`

If `f''(c) > 0` we have a relative minimum

If `f''(c) < 0` we have a relative maximum

`f''(4/3 +(sqrt(13))/3) > 0` relative min at `x = 4/3 +(sqrt(13))/3`

`f''(4/3 - (sqrt(13))/3) < 0` relative max at `x = 4/3 - (sqrt(13))/3`