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#.
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Answer 2

To find the critical points for (y = x^2 \cdot 2^x), you first take the derivative of the function, set it equal to zero, and then solve for (x). These points correspond to potential maxima or minima.

Given (y = x^2 \cdot 2^x), its derivative is (y' = 2x \cdot 2^x + x^2 \cdot \ln(2) \cdot 2^x).

Setting (y' = 0): [2x \cdot 2^x + x^2 \cdot \ln(2) \cdot 2^x = 0] [2x \cdot 2^x + x^2 \cdot \ln(2) \cdot 2^x = 2^x(2x + x^2 \cdot \ln(2)) = 0] This equation equals zero when (2^x = 0) (which is not possible) or when (2x + x^2 \cdot \ln(2) = 0).

Solving (2x + x^2 \cdot \ln(2) = 0): [x(2 + x \cdot \ln(2)) = 0] So, (x = 0) or (2 + x \cdot \ln(2) = 0).

When (x = 0), (y = 0).

For (2 + x \cdot \ln(2) = 0): [x \cdot \ln(2) = -2] [x = \frac{-2}{\ln(2)}]

Thus, the critical points are (x = 0) and (x = \frac{-2}{\ln(2)}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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