How do you find the exact relative maximum and minimum of the polynomial function of # f(x) = –2x^3 + 6x^2 + 18x –18 #?

Question

Isaiah Allen · Accepted Answer

The relative maximum is #=(3,36)# and the relative minimum is #=(-1,-28)#

The polynomial function is

#f(x)=-2x^3+6x^2+18x-18#

This function is defined, continuous and derivable on #RR#

The first derivative is

#f'(x)=-6x^2+12x+18#

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

#-6(x^2-2x-3)=0#

#(x+1)(x-3)=0#

#f'(x)=-(x+1)(x-3)#

Therefore,

#x=-1# and #x=3#

We can make a variation chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaa)##3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+1##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-3##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f'(x)##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaa)##↘##color(white)(aaaa)##↗##color(white)(aaa)##↘#

The second derivative is

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

#f''(-1)=12+12=24#, #=>#, this is a relative minimum as #f''(-1)>0#

#f''(3)=-12*3+12=-24#, #=>#, this is a relative maximum as #f''(-1)<0#

graph{-2x^3+6x^2+18x-18 [-74.1, 74.1, -37.03, 37]}

Aurora Alvarado · Answer

undefined To find the exact relative maximum and minimum of the polynomial function $ f(x) = -2x^3 + 6x^2 + 18x - 18 $, you first need to find its critical points by taking the derivative and setting it equal to zero. Then, you can use the second derivative test to determine whether each critical point corresponds to a relative maximum, minimum, or neither.

1. Find the derivative of $ f(x) $:
$$ f'(x) = -6x^2 + 12x + 18 $$

2. Set $ f'(x) $ equal to zero and solve for $ x $:
$$ -6x^2 + 12x + 18 = 0 $$
$$ -2x^2 + 4x + 6 = 0 $$
$$ -x^2 + 2x + 3 = 0 $$
$$ (x - 3)(-x - 1) = 0 $$

The critical points are $ x = 3 $ and $ x = -1 $.

3. Find the second derivative $ f''(x) $:
$$ f''(x) = -12x + 12 $$

4. Evaluate $ f''(x) $ at the critical points:
$$ f''(3) = -12(3) + 12 = -36 + 12 = -24 $$ (negative, so it's a local maximum)
$$ f''(-1) = -12(-1) + 12 = 12 + 12 = 24 $$ (positive, so it's a local minimum)

5. Therefore, the function has a relative maximum at $ x = 3 $ and a relative minimum at $ x = -1 $.