Let f : R rightarrow R be defined by f(x) = x | vedclass

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(A) Given the function $f(x) = x^3 - 3x^2 + 3x - 2$. We can rewrite the function as $f(x) = (x^3 - 3x^2 + 3x - 1) - 1$. This simplifies to $f(x) = (x

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$1 + \sqrt[3]{x + 1}$

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(A) Given the function $f(x) = x^3 - 3x^2 + 3x - 2$. We can rewrite the function as $f(x) = (x^3 - 3x^2 + 3x - 1) - 1$. This simplifies to $f(x) = (x - 1)^3 - 1$. To find the inverse $f^{-1}(x)$,let $y = f(x)$. So,$y = (x - 1)^3 - 1$. Add $1$ to both sides: $y + 1 = (x - 1)^3$. Take the cube root of both sides: $\sqrt[3]{y + 1} = x - 1$. Add $1$ to both sides: $x = 1 + \sqrt[3]{y + 1}$. Replacing $y$ with $x$,we get $f^{-1}(x) = 1 + \sqrt[3]{x + 1}$.

Let $f : R \rightarrow R$ be defined by $f(x) = x^3 - 3x^2 + 3x - 2$. Then $f^{-1}(x)$ is given by:

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