f: (-infty, 0] rightarrow [0, infty) is defin | vedclass

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(A) We have a function $f: (-\infty, 0] \rightarrow [0, \infty)$ defined as $f(x) = x^2$.Since each line parallel to the $x$-axis cuts the curve at mo

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Domain of $f^{-1} = [0, \infty)$,Range of $f^{-1} = (-\infty, 0]$

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(A) We have a function $f: (-\infty, 0] \rightarrow [0, \infty)$ defined as $f(x) = x^2$.Since each line parallel to the $x$-axis cuts the curve at most at one point,the function $f$ is one-one.From the graph,it is clear that the range of $f$ is $[0, \infty)$,which is equal to the codomain.Therefore,$f$ is onto.Since $f$ is both one-one and onto,it is invertible.The inverse function $f^{-1}$ maps the codomain of $f$ to the domain of $f$.Thus,$f^{-1}: [0, \infty) \rightarrow (-\infty, 0]$.Hence,the domain of $f^{-1}$ is $[0, \infty)$ and the range of $f^{-1}$ is $(-\infty, 0]$.

$f: (-\infty, 0] \rightarrow [0, \infty)$ is defined as $f(x) = x^2$. The domain and range of its inverse are

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