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Find the range of this function $h(x)= {(3x+2)(x-2)}/{x(x-2)}$?

I just don't know how to find the answer. It is $y \neq 3$ But how??

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The domain of the function is $(-\infty, 0) \cup (0,2) \cup (2, \infty)$.

At each interval find out if the the function is increasing or decreasing.

If the function is incresing at an interval $(a,b)$ the range is $(\lim_{x \rightarrow a}f(x), \lim_{x \rightarrow b} f(x))$.

If the function is decreasing at an interval $(a,b)$ the range is $(\lim_{x \rightarrow b}f(x), \lim_{x \rightarrow a} f(x))$.

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Use contradiction. First, multiply to get ${(3x^2-4x-4)}/{(x^2-2x)}$. Then set $y=3$. By cross multiplying, you will see that $3x^2-6x=3x^2-4x-4$. If you simplify, you will get $2x=4$ or $x=2$. If you plug this in, you will see that the bottom of the fraction equals 0, so $y=3$ doesn't work.

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Clearly, the function has a removable discontinuity at $x=2$. Removing that, we have: $$h(x)=\frac{3x+2}{x}$$ Which has a vertical asymptote at $x=0$. Now, what are the horizontal asymptotes? You should find that they are both equal to $y=3$. Now, does the function cross the asymptotes? Show the function has no local extrema and then plot the function.

You can also see this by solving for $x$: $$xh(x)=3x+2\ \longrightarrow \ x(h(x)-3)=2$$ Clearly, when $x=0$ or $h(x)=3$ there is a contradiction, so we know that $h(x)\neq 3$.

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Assuming you mean that $h(x)=\frac{(3x+2)(x-2)}{x(x-2)}$, the function is the same as $\frac{3x+2}{x}$, which is $3+\frac2x$ except at $x=2$, where $y$ would otherwise be $4$. Such a function is a transform of $y=\frac1x$, which is one to one and only has one missing point from its domain, which is the limit of the function as $x\to\infty$. In this case, that is $3$. So both $3$ and $4$ are missing from the range. The range is $(-\infty,3)\cup(3,4)\cup(4\infty)$.

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