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

1 Answer
Jul 21, 2018

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#