How do you find the exact relative maximum and minimum of the polynomial function of #f(x) = x^3 - 2x^2 - x +1#?

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Answer 1 · 2016-02-17T14:03:38

Find and test the critical numbers, then find #f(c)# for them.

#f(x) = x^3 - 2x^2 - x +1#

#f'(x)=3x^2-4x-1# which is never undefined and is #0# at

#x = (4+-sqrt(16+12))/6 = (2+-sqrt7)/3#

The critical numbers are #(2-sqrt7)/3# and #(2+sqrt7)/3#.

The sign of #f'(x)# is positive a little left of #(2-sqrt7)/3# and negative a little to the right.
Therefore, #f((2-sqrt7)/3)# is a relative maximum.

#f((2-sqrt7)/3) = ((2-sqrt7)/3)^3 - 2((2-sqrt7)/3)^2 - ((2-sqrt7)/3) +1#

# = 7/(3sqrt3)(2sqrt7-1)#

The sign of #f'(x)# is negative a little left of #(2+sqrt7)/3# and positive to the right.
Therefore, #f((2+sqrt7)/3)# is a relative minimum.

#f((2+sqrt7)/3) = ((2+sqrt7)/3)^3 - 2((2+sqrt7)/3)^2 - ((2+sqrt7)/3) +1#

# = -7/(3sqrt3)(2sqrt7+1)#