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Let \( f: R \rightarrow R \) be defined by \( f(x)=2 x+6 \) which is a bijective mapping then \( f^{-1}(x) \) is given by
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The correct answer is:
\( \frac{x}{2}-3 \)
Given that $f(x)=2 x+6, f: R \rightarrow R$
Let $y=2 x+6$ then, $2 x=y-6$
$\Rightarrow x=\frac{y-6}{2}$
Now, $y=\frac{x-6}{2}$
$=\frac{x}{2}-3$
Therefore, $f^{-1}(x)=\frac{x}{2}-3$
Let $y=2 x+6$ then, $2 x=y-6$
$\Rightarrow x=\frac{y-6}{2}$
Now, $y=\frac{x-6}{2}$
$=\frac{x}{2}-3$
Therefore, $f^{-1}(x)=\frac{x}{2}-3$
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