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Q.
If $f : R \to R$ be a mapping defined by $f(x) = x^3 + 5$, then $f^{-1}(x)$ is equal to:
Relations and Functions - Part 2
Solution:
Let f : R $\to$ R defined as
$f (x) = x^3 + 5$
Let f (x) = y $\Rightarrow $ x = $f^{-1}$(y) .....(i)
Given : $f(x) = x^3 + 5$
$\Rightarrow \, y = x^3 + 5$
$\Rightarrow \, y - 5 = x^3$
$\Rightarrow \, (y - 5)^{\frac{1}{3}} = x = f^{-1} (y)$ (from (i))
$\therefore \, f^{-1} (x) = (x - 5)^{\frac{1}{3}}$