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What is the range the function $h(x)=\frac{x-2}{x+3}$ ?
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2835 Upvotes
Verified Answer
The correct answer is:
$(-\infty, 1) \cup(1, \infty)$
To find, range of the function $h(x)=\frac{x-2}{x+3}$
Let $h(x)=y$
i.e. set of possible values of $y$ will give the range of $h(x)$.
$$
\begin{aligned}
& \Rightarrow & \frac{x-2}{x+3} & =y \\
& & x-2 & =x y+3 y \\
\Rightarrow & & x-x y & =3 y+2 \\
\Rightarrow & & x & =\frac{3 y+2}{1-y}
\end{aligned}
$$
which is not defined for $y=1$
Hence, range $=R-\{1\}$ or $(-\infty,-1) \cup(1, \infty)$
Let $h(x)=y$
i.e. set of possible values of $y$ will give the range of $h(x)$.
$$
\begin{aligned}
& \Rightarrow & \frac{x-2}{x+3} & =y \\
& & x-2 & =x y+3 y \\
\Rightarrow & & x-x y & =3 y+2 \\
\Rightarrow & & x & =\frac{3 y+2}{1-y}
\end{aligned}
$$
which is not defined for $y=1$
Hence, range $=R-\{1\}$ or $(-\infty,-1) \cup(1, \infty)$
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