2x^3+3x^2-120x

to find the critical numbers I took the derivative which is 6x^2+6x-120 so divide by 6 to simplify

x^2+x-20=0

so the critical numbers will be (4,0) and (-5,0) but its wrong I don't know y?

Your work looks good to me, but the critical numbers are 4 and -5, not the points you give.

Recall the definition:

A number a in the domain of a given function f is called a critical number of f if f '(a) = 0 or f' is undefined at x = a.

Thnks :)

To find the critical numbers of a function, you need to find the values of x where the derivative of the function is equal to zero or undefined.

In your case, you correctly took the derivative of the function 2x^3+3x^2-120x, which is 6x^2+6x-120. However, there seems to be an error when simplifying the derived expression.

Let's correct the error by dividing the derived expression by 6:

(6x^2+6x-120) / 6 = x^2 + x - 20

Now, this is a quadratic equation. To find its critical numbers, we need to find the values of x where the equation equals zero:

x^2 + x - 20 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, let's factor it:

(x+5)(x-4) = 0

Setting each factor to zero:

x + 5 = 0 --> x = -5
x - 4 = 0 --> x = 4

So, the critical numbers of the function 2x^3+3x^2-120x are x = -5 and x = 4.

If your answer is different from this, please check the steps you followed or recheck your calculations to identify any errors.