Question

5b) find the minimum and maximum function of f(x) = x3+two x square -4x+6

Answer 1

f( x ) = x³ +2x² -4x +6

differentiate with respect to x

df(x)/dx = 3x² +4x -4 =0

3x² +4x -4 = 0
3x² +6x -2x -4 =0
3x(x +2) -2(x +2) =0
(3x -2)(x +2) =0
x = -2 , 2/3

if you put x = -2 and 2/3 in number line
you see at x = -2
dy/dx change sign positive to negative
hence, at x = -2 function gain maximum value

so, maximum value of function, is
= (-2)³+2(-2)²-4(-2)+6
=-8 +8 +8 +6
=14

similarly at x = 2/3 function gain minimum value ,

minimum value of f(x) = (2/3)³ +2(2/3)²-4(2/3)+6

=8/27 +8/9 -8/3 + 6

=8{ 1/27 +1/9 -1/3} + 6
=-40/27 +6
= (162-40)/27
=122/27