Global maximum question

Question

The global maximum over the indicated interval for y = x^3 - 3x^2 - 3x + 6, -2≤x≤4 is (_,_). Type your answer to 2 decimal places.

Answer 1

y= x^3 -3x^2 -3x +6

y' = 3x^2 -6x -3 = 0

x^2 -2x -1=0

x=1+/-.5sqr8= 1+/-sqr2= about 2.414, -.414

y"= 6x-6

y"(1+sqr2)= sqr2>0, concave up, y(1+sqr2)= local min

y"(1-sqr2)=-sqr2<0, concave down, y(1-sqr2) = local max

check the endpoints

y(-2)= -8-12+6+6=-8= global minimum

y(4)=64-48-12+6=10= global maximum

y(1+sqr2)=(1+sqr2)^3-3(1+sqr2)^2 -3(1+sqr2)+6

=(3+2sqr2)(1+sqr2)-9-6sqr2-3-3sqr2 +6

= 7+5sqr2-12-9sqr2 +6

=1-4sqr2

=about 1-4(1.414)

= about 1-4.242

=about -3.242 = local minimum

y(x)= x^3-3x^2-3x+6

for x=1-sqr2

y(1-sqr2)= (1-sqr2)^3-3(1-sqr2)^2-3(1-sqr2)+6

=(3-2sqr2)(1-sqr2)-9+6sqr2-3+3sqr2 +6

=7-5sqr2 -12 +9sqr2+6

= 1+4sqr2

= about 1+4(1.414)

= about 1+5.656

=about 6.656 = local maximum for interval x=-2 to 4

Answer 2

Take derivative,set to zero to get the extrema (max.or min.points)

both values are in range (~-0.414,2.414)

The values of the function at these points are:

The first value is clearly a minimum point and the second value is clearly a maximum point.

The answer is the second value.