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

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Answer 1 · 2015-03-30T14:15:50

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 $x=-\frac{2}{ln(2)}\approx -2.885$ .

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