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Consider $f: R^{+} \rightarrow[4, \infty)$ given by $f(x)=x^2+4$. Show that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}(y)$ $=\sqrt{y-4}$, where $\mathrm{R}^{+}$is the set of all non-negative real numbers
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$f\left(x_1\right)=x_1^2+4$ and $f\left(x_2\right)=x_2^2+4$
$f\left(x_1\right)=f\left(x_2\right) \Rightarrow x_1^2+4=x_2^2+4$
or $\mathrm{x}_1^2=\mathrm{x}_2^2 \Rightarrow \mathrm{x}_1=\mathrm{x}_2 \quad$ As $\mathrm{x} \in \mathrm{R}$
$\therefore \mathrm{x}>0, \mathrm{x}_1{ }^2=\mathrm{x}_2{ }^2 \Rightarrow \mathrm{x}_1=\mathrm{x}_2 \Rightarrow \mathrm{f}$ is one-one
Let $y=x^2+4$ or $x^2=y-4$ or $x=\pm \sqrt{y-4}$
$x$ being $>0,-v e$ sign not to be taken
$$
\therefore x=\sqrt{y-4} \therefore f^{-1}(y)=g(y)=\sqrt{y-4}, y \geq 4
$$
For every $\mathrm{y} \geq 4, \mathrm{~g}(\mathrm{y})$ has real positive value.
$\therefore$ The inverse of $f$ is $\mathrm{f}^{-1}(\mathrm{y})=\sqrt{\mathrm{y-4}}$
$f\left(x_1\right)=f\left(x_2\right) \Rightarrow x_1^2+4=x_2^2+4$
or $\mathrm{x}_1^2=\mathrm{x}_2^2 \Rightarrow \mathrm{x}_1=\mathrm{x}_2 \quad$ As $\mathrm{x} \in \mathrm{R}$
$\therefore \mathrm{x}>0, \mathrm{x}_1{ }^2=\mathrm{x}_2{ }^2 \Rightarrow \mathrm{x}_1=\mathrm{x}_2 \Rightarrow \mathrm{f}$ is one-one
Let $y=x^2+4$ or $x^2=y-4$ or $x=\pm \sqrt{y-4}$
$x$ being $>0,-v e$ sign not to be taken
$$
\therefore x=\sqrt{y-4} \therefore f^{-1}(y)=g(y)=\sqrt{y-4}, y \geq 4
$$
For every $\mathrm{y} \geq 4, \mathrm{~g}(\mathrm{y})$ has real positive value.
$\therefore$ The inverse of $f$ is $\mathrm{f}^{-1}(\mathrm{y})=\sqrt{\mathrm{y-4}}$
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