How do you find the critical points and the open intervals where the function is increasing and decreasing for y = xe^(x(2 - 3x))?

Jun 17, 2015

The function is increasing on the interval: $\left(\frac{1 - \sqrt{7}}{6} , \frac{1 + \sqrt{7}}{6}\right)$ and decreasing on the intervals: $\left(- \infty , \frac{1 - \sqrt{7}}{6}\right)$ and $\left(\frac{1 + \sqrt{7}}{6} , \infty\right)$

Explanation:

Jasivan S, gives an excellent answer to the question asked, but I suspect that the intended question was: Find the open intervals on which the function is increasing and those on which it is decreasing. (Rather than finding those on which it is doing both.)

See Jasivan S. answer to find the derivative: $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(- 6 {x}^{2} + 2 x + 1\right) {e}^{2 x - 3 {x}^{2}}$

and the critical numbers: ${x}_{1} = \frac{1 + \sqrt{7}}{6}$ and ${x}_{2} = \frac{1 - \sqrt{7}}{6}$.

(I'm used to thinking of critical points as points in the domain, so these would be the critical points. Some think of critical points as points on a graph, in which case, you'd need the $y$ values as well. See Jasivan S. answer for that.)

As you probably know, to investigate increasing and decreasing behavior, we consider the sign of $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Because e^23-3x^2) is always positive, the sign of $\frac{\mathrm{dy}}{\mathrm{dx}}$ will be the same as the sign of $- 6 {x}^{2} + 2 x + 1$ which is
positive on $\left(\frac{1 - \sqrt{7}}{6} , \frac{1 + \sqrt{7}}{6}\right)$, and
negative on $\left(- \infty , \frac{1 - \sqrt{7}}{6}\right)$ and on $\left(\frac{1 + \sqrt{7}}{6} , \infty\right)$.

Cosequently,

The function is increasing on the interval: $\left(\frac{1 - \sqrt{7}}{6} , \frac{1 + \sqrt{7}}{6}\right)$ and decreasing on the intervals: $\left(- \infty , \frac{1 - \sqrt{7}}{6}\right)$ and $\left(\frac{1 + \sqrt{7}}{6} , \infty\right)$