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

Mar 25, 2018

Determine the first derivative:

$y ' = \left(12 {x}^{3} - 24 {x}^{2} - 36 x\right) {e}^{3 {x}^{4} - 8 {x}^{3} - 18 {x}^{2}}$

We need to determine the candidates for maximum/minimum. These are critical points, and occur when $y ' = 0$. We can immediately get rid of ${e}^{3 {x}^{4} - 8 {x}^{3} - 18 {x}^{2}}$ because this will never equal $0$.

$0 = 12 {x}^{3} - 24 {x}^{2} - 36 x$

$0 = {x}^{3} - 2 {x}^{2} - 3 x$

$0 = x \left({x}^{2} - 2 x - 3\right)$

$0 = x \left(x - 3\right) \left(x + 1\right)$

$x = 0 , 3 \mathmr{and} - 1$

Now we need to test the value of the derivative within these intervals to determine the nature of these points. We need not include ${e}^{3 {x}^{4} - 8 {x}^{3} - 18 {x}^{2}}$ here because this will never influence the sign of the derivative.

We see that

$y ' \left(1\right) = 12 {\left(1\right)}^{3} - 24 {\left(1\right)}^{2} - 36 \left(1\right) = \text{negative}$

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!