# How do you find the critical points for y = x^2(2^x)?

Use the Product Rule to get $\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \setminus \cdot {2}^{x} + {x}^{2} \setminus \cdot \ln \left(2\right) \setminus \cdot {2}^{x} = {2}^{x} \left(2 x + \ln \left(2\right) {x}^{2}\right) = x {2}^{x} \left(2 + \ln \left(2\right) x\right)$.
Setting this equal to zero and solving for $x$ gives critical points at $x = 0$ and $x = - \setminus \frac{2}{\ln \left(2\right)} \setminus \approx - 2.885$.