What is 2xe^x+e^xx^2>0 ?

The solution to 2xe^x+e^xx^2>0 is x 0

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2xe^x+e^xx^2>0

2 x e^{x} + e^{x} x^{2} > 0

x < - 2 or x > 0

Interval Notation

(- \infty, - 2) \cup (0, \infty)

Solution steps

2 x e^{x} + e^{x} x^{2} > 0

Apply exponent rules

2 x e^{x} + e^{x} xx

Factor out common term

x e^{x}

x e^{x} (2 + x)

Find the signs of the factors of

e^{x} x (2 + x)

Identify the intervals:

x < - 2, x = - 2, - 2 < x < 0, x = 0, x > 0

Summarize in a table:

e^{x} x 2 + x e^{x} x (2 + x) x < - 2 + - - + x = - 2 + - 00 - 2 < x < 0 + - + - x = 0 + 0 + 0 x > 0 + + + +

Identify the intervals that satisfy the required condition:

> 0

x < - 2 or x > 0

2. 4^{2}

P (q, s) = q + 2 + s

\frac{0 ^{3}}{3}

0.2 - \frac{4}{10} - 1.625

\frac{a ^{\frac{3}{4}} b ^{- 1} \cdot b ^{\frac{7}{4}}}{3 b ^{- 1}}