Find range of f(x) = `1/sqrt(16- 4^(x^2-x))`

Explanation

Here, `f(x) = 1/sqrt(16-4^(x^2-x)`

We know, square root of a value is greater than `0`,

`:. sqrt(16-4^(x^2-x) gt 0`

`=>(16-4^(x^2-x) gt 0`

`=>4^2 gt 4^(x^2-x)`

`=>x^2-x lt 2`

`=>x^2-x-2 lt 0`

`=>(x-2)(x+1) lt 0`

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