Question

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

Answer 1

Use the Product Rule to get `\frac{dy}{dx}=2x\cdot 2^{x}+x^{2}\cdot ln(2)\cdot 2^{x}=2^{x}(2x+ln(2)x^{2})=x2^{x}(2+ln(2)x)`.

Setting this equal to zero and solving for `x` gives critical points at `x=0` and `x=-\frac{2}{ln(2)}\approx -2.885`.