$(1)$If $f(x)=x^n$ and $g(x)=x^\frac{1}{n} $,then $f(x)$ and $g(x)$ are the inverse of each other.
$(2)$ therefore $f(x) = x^n$ and $g(x) = x^\frac{1}{n}$ is the mirror image about $y=x$ Now how $(2)$ follows from $(1)$
How to prove it?
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See https://math.stackexchange.com/questions/387542/is-there-an-explanation-why-the-reflection-of-fx-through-y-x-is-its-invers – A. Goodier Jun 03 '18 at 10:06
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@A.Goodier ok thank you, sir. – Tanvi Sorout Jun 03 '18 at 10:09
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The statement that $f(x) = x^n$ and $g(x) = x^{1/n}$ are inverses of each other is only true when $n$ is odd unless you restrict the domain of $f$ to the nonnegative real numbers. – N. F. Taussig Jun 03 '18 at 10:09
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Note that for every point $(a,b)$ on $f$ there is a point $(b,a)$ on $g$. This explains the $y=x$ reflection and why all points of equivalence of $f=g$ are on $y=x$ – Rhys Hughes Jun 03 '18 at 10:19