The maximum and minimum measurements are 5 and -3 respectively.
f ( x ) = -x^3 +3x^2+ 45 x + 10
f' l x ) = 0
0 = - 3x^2 +6x +45
x = 5,- 3.
f ( 4 ) = - ( 4 )^3+ 3( 4 )^ 2 + 45 ( 4 ) + 10 = 174
f ( - 4 ) = - ( - 4 )^3 + 3( - 4 )^2 + 45(- 4) + 10 = - 58
f ( - 3 ) = -(- 3 )^3 + 3 ( - 3)^2 + 45 (- 3) + 10 = - 7 1
+ ( 5 ) = -(5)^3 + 3(5 )^2 +45(5)+ 10 = 185
Absolute maximum = maxis{ f(4 ),f(5),f(-4),f( - 3 )}
= 185 at x=5
Absolute mininum=minis{f(4 ),f(5),f(-4),f( - 3 )}
= - 71 at x= -3
In light of this, the maximum and minimum values are 5 and -3, respectively.
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