Question

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

Answer 1

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 `y' = 0`. We can immediately get rid of `e^(3x^4 - 8x^3 - 18x^2)` because this will never equal `0`.

`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 `e^(3x^4 - 8x^3 - 18x^2)` here because this will never influence the sign of the derivative.

We see that

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

Therefore, `x = 0` is a local maximum, `x = 3` is a local minimum and `x = -1` is a local minimum.

We can confirm this graphically.

Hopefully this helps!