For the function h(x) =(x+2) ^3-3x-2,
Find the local minimum and maximum.
First find the derivative using the chain rule then h'(x) = 3(x+2)2 - 3 . When the derivative of the function is set to zero, this means that the slope of the tangent to the function curve is 0, which means that the slope is flat at that point, and the only time you will have a flat slope at the curve is when you are at either a minimum or a maximum.
h'(x) = 3(x+2)2 - 3 = 0, 3(x+2)2 = 3 , (x+2)2 = 1 , take square root of both sides ( x+2) = ±1
(x+2) = 1, and ( x+2) = -1, x = -1, and x = -3. The points -1 and -3 are two possible local maximum and minimum but we have to examine them first.
1- for x = -3 examine the slope before and after the point
Choose then points x = -4 , and x = -2
h'(-4) = 3(-4+2)2 - 3 = 9 >0 is a positive slope just before the point x = -3
h'(-2) = 3(-2+2)2 - 3 = -3 < 0 is a negative slope just after the point x= -3
Therefore x = -3 is a relative maximum
1- for x = -1 examine the slope before and after the point
Choose the points x = -2 , and x = 0
h'(-2) = 3(-2+2)2 - 3 = -3 < 0 is a negative slope just before the point x = -3
h'(0) = 3(0+2)2 - 3 = 9 > 0 is a positive slope just after the point x= -3
Therefore x = -1 is a relative minimum
