How do you find the exact relative maximum and minimum of the polynomial function of #f(x)=2x^3+x^2-11x#?

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Answer 1 · 2017-08-09T15:41:32

The relative minimum is #=(1.198,-8.304)# and the relative maximum is #=(-1.531, 12.008)#

The function is

#f(x)=2x^3+x^2-11x#

We calculate the first derivative

#f'(x)=6x^2+2x-11#

The critical points are when #f'(x)=0#

#6x^2+2x-11=0#

We solve this like a quadratic equation, we start by calculating the discriminant

#Delta=b^2-4ac=2^2-4(6)(-11)=268#

As #Delta >0#, there are #2# real roots

The solutions are

#x=(-b+-sqrtDelta)/(2a)=(-2+-sqrt268)/(12)#

#x_1=(-2-sqrt268)/(12)=-1.531#, #=>#, #y_1=12.008#

#x_2=(-2+sqrt268)/(12)=1.198#, #=>#, #y_2=-8.304#

We calculate the second derivative

#f''(x)=12x+2#

Therefore,

#f''(x_1)=12*-1.531+2=-16.372<0#, #=>#, this is a relative maximum

#f''(x_2)=12*1.198+2=16.67>0#, #=>#, this is a relative minimum

graph{2x^3+x^2-11x [-30.5, 27.21, -10.8, 18.08]}