Question

The maximum and minimum values of the function y=x^3- 3x^2+6
1) 2,0
2) 6,0
3) 6,2
4) 4,2

Answer 1

Answer:

a is correct answer

Explanation:

Given, Function,

y=x³-3x²+6

So for that, we have to find the critical value of y so, differentiate y respect to x

y'=3x²-6x

=3x(x-2) =0

So, X is either be 2 or 0. these are the critical values.