# How do you find the critical numbers for #y = e^(3x^4-8x^3-18x^2 )# to determine the maximum and minimum?

##### 1 Answer

Determine the first derivative:

#y' = (12x^3 - 24x^2 - 36x)e^(3x^4 - 8x^3 - 18x^2)#

We need to determine the candidates for maximum/minimum. These are critical points, and occur when

#0 = 12x^3 - 24x^2 - 36x#

#0 = x^3 - 2x^2 - 3x#

#0 = x(x^2 -2x - 3)#

#0 = x(x - 3)(x + 1)#

#x= 0, 3 or -1#

Now we need to test the value of the derivative within these intervals to determine the nature of these points. We need not include

We see that

#y'(1) = 12(1)^3 - 24(1)^2 - 36(1) = "negative"#

Therefore,

We can confirm this graphically.

Hopefully this helps!