If f(x) = 3x - 5,then {f^{ - 1}}(x) is: | vedclass

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(B) Let $f(x) = y$. Since $f(x) = 3x - 5$,we have $y = 3x - 5$. To find the inverse,we solve for $x$ in terms of $y$: $y + 5 = 3x$ $x = \frac{y + 5}{3

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Given by $\frac{{x + 5}}{3}$

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(B) Let $f(x) = y$. Since $f(x) = 3x - 5$,we have $y = 3x - 5$. To find the inverse,we solve for $x$ in terms of $y$: $y + 5 = 3x$ $x = \frac{y + 5}{3}$. By definition,$x = f^{-1}(y)$,so $f^{-1}(y) = \frac{y + 5}{3}$. Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x + 5}{3}$. Since $f(x) = 3x - 5$ is a linear function,it is both one-one (injective) and onto (surjective),therefore $f$ is invertible.

If $f(x) = 3x - 5$,then ${f^{ - 1}}(x)$ is:

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