Question

Maximum and Minimum values of the function y=x³-3x²+6 are​

Answer 1

Step-by-step explanation:

Differentiating the equation:

dy/dx =3x^2–6x

We have to now take the values of x by:

3x^2–6x=0

x(3x-6)=0

Which means:

x=0 or x=2

So, we get two cases.

Now, double differentiate the equation to find whether the equation is maximum or minimum at that point.

If, d^2(y)/(dx)^2=6x-6 is <0 then the point gives maximum value and vice versa.

Now, apply the cases:

Case 1: at x=0

d^2(y)/(dx)^2= -6.

So, the point x=0; gives maximum value

Case 2: at x=2

d^2(y)/(dx)^2=6.

So, the point gives Minimum value at x=2.

Hence, the maximum and minimum values are:

Maximum value = (0)^3–3(0)^2+6 = 6

And the minimum value = (2)^3–3(2)^2+6 = 2.

Answer 2

Answer:

check below ( not my answer but wanted to help! )

Step-by-step explanation:

Given, Function,y= x^ 3  − 3x^2 + 6

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

y   =3x^2  −6x = 3x(x−2)=0

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