find the maximum and minimum;2X^3-3X^2-12X+18

Question

hi am talo please help me in this question

Answer 1

Hello Talo,

 

The first step to find the min/max values is to find the derivative.

 

y=2x³-3x²-12x+18

y'=6x²-6x-12

 

Then we find the x intercepts of the derivative.

0=6x²-6x-12

0=6(x²-x-2)

0=6(x-2)(x+1)

x=2,-1

 

The values for the min/max are x=-1 and x=2.  This is because these are the values at which the derivative is 0.  Since the derivative tells the instantaneous slope of the function at any point, 0 values indicate that the slope is changing from positive to negative, which occurs at mix/max values.

 

We solve for these points in the original function:

y(-1)=-2-3+12+18=31      (-1,25)

y(2)=16-12-24+18=-10     (2,-10)

 

However to determine which is min and which is max, we need the 2nd derivative.

 

y'=6x²-6x-12

y''=12x-6

 

Then we plug in the values we found in the last step.

y''(-1)=12(-1)-6=-18

y''(2)=12(2)-6=18

 

This is all the info we needed.  Since y''(-1) is negative this tells us that this point is a maximum, since a negative 2nd derivative tells us that the slope is decreasing.

Y''(2) is positive, so this is a minimum value since this shows that the slope is increasing.

 

Bottom line:

Min (2,-10)

Max (-1,25)

Answer 2

All you need to do is differentiate wrt x and equate to zero. The solutions are either maxima or minima.

 

Given expression = 2x^3 - 3x^2 - 12x +18

 

Differential wrt x = 6x^2 - 6x - 12 = 0       At the extrema points, the slopes of the function are zero.

 

The solutions are x=2 and x=-1 each of which are either maxima or minima.

 

To find out which is maximum and which is minimum, differntiate again wrt x

 

2nd Differential = 12x-6.

 

Now at x=2, 12x-6 is positive, so the minimum is here = 2(8)-3(4)-12(2)+18 = -2

 

At x=-1, 12x-6 is negative, so the maximum is here = 2(-1)-3(1)-12(-1)+18 = 25