What is the extreme f(x)=e^{1/x}*(x+2) ?

The extreme f(x)=e^{1/x}*(x+2) is Maximum(-1, 1/e),Minimum(2,4e^{1/2})

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extreme f(x)=e^{1/x}*(x+2)

extreme

Maximum (- 1, \frac{1}{e}), Minimum (2, 4 e^{\frac{1}{2}})

Solution steps

Find the critical points:

x = - 1, x = 2

Domain of

e^{\frac{1}{x}} \cdot (x + 2) : x < 0 or x > 0

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = - 1

into

e^{\frac{1}{x}} \cdot (x + 2) \Rightarrow y = \frac{1}{e}

Maximum (- 1, \frac{1}{e})

Plug the extreme point

x = 2

into

e^{\frac{1}{x}} \cdot (x + 2) \Rightarrow y = 4 e^{\frac{1}{2}}

Minimum (2, 4 e^{\frac{1}{2}})

Maximum (- 1, \frac{1}{e}), Minimum (2, 4 e^{\frac{1}{2}})

What is the extreme f(x)=e^{1/x}*(x+2) ?
The extreme f(x)=e^{1/x}*(x+2) is Maximum(-1, 1/e),Minimum(2,4e^{1/2})