Consider the function f(x)=xex+1xex where x∈R−{0}, then which of the following statement is CORRECT?

f ( x ) attains its local minima at x = x 0 , where x 0 ∈ ( 1 , ∞ )

f ( x ) attains its local minima at x = x 0 , where x 0 ∈ ( 1 , ∞ )

f ( x ) attains its local minima at x = x 0 , where x 0 ∈ ( 0 , 1 )

f ( x ) attains its local maxima at x = x 0 , where x 0 ∈ ( 0 , 1 )

f ( x ) attains its local minima at x = x 0 , where x 0 ∈ ( 1 , ∞ )

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Consider the function $f (x) = x e^{x} + \frac{1}{x e^{x}}$ where $x \in R - {0},$ then which of the following statement is CORRECT?

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$f (x)$ attains its local minima at $x = - 1$

$f (x)$ attains its local minima at $x = x_{0},$ where $x_{0} \in (0, 1)$

$f (x)$ attains its local maxima at $x = x_{0},$ where $x_{0} \in (0, 1)$

$f (x)$ attains its local minima at $x = x_{0},$ where $x_{0} \in (1, \infty)$

answer is D.

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$f (x) = x e^{x} + \frac{1}{x e^{x}}$ where $x \neq 0$

$f' (x) = \frac{e^{3 x} (x + 1) (x + e^{- x}) (x - e^{- x})}{{(x e^{x})}^{2}}$

$f' (x) = 0 \Rightarrow x = - 1; e^{- x} = - x$ (no solution)

$e^{- x} = x$ (exactly one solution)

At $x = - 1,$ local maxima.

At $x = x_{0} \in (0, 1)$ local minima.

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