Question

4.1.
У=4x^3-6x^2+12x-10.
findout maxima minima?​

Answer 1

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