Show that `f(x)=e^(1/x),x!=0` is decreasing function for all `x!=0.`

Explanation

`f(x)=e^{frac{1}{x}} `
`
f^{prime}(x)=e^{frac{1}{x}} frac{d}{d x}(frac{1}{x}) `
`
=e^{frac{1}{x}}(frac{-1}{x^{2}}) `
`
=-frac{e^{frac{1}{x}}}{x^{2}}`


Here,` e^{frac{1}{x}}>0 and x^{2}>0, for any real value of x neq 0. `


Therefore,` f(x)=-frac{e^{frac{1}{x}}}{x^{2}}<0, forall x in R, x neq 0`

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