How to find the range of a function without a graph?
Okay so I know how to find the domain of the function $f(x) = \frac{3x^2}{x^2-1}$, which is $x\neq 1$ and $x\neq -1$, but I'm totally confused on how to find the range without using a graph.
$\endgroup$ 43 Answers
$\begingroup$Set $$y=\frac{3x^2}{x^2-1}$$ and rearrange to get $$(y-3)x^2-y=0$$ This quadratic has real roots provided $b^2-4ac\geq 0$ which translates as $$y(y-3)\geq 0$$ Noting that $y=3$ is the horizontal asymptote, the solution set, i.e.the range, is $$y\leq 0, y>3$$
$\endgroup$ $\begingroup$Write the function as $$f(x)=\frac{3x^2-3+3}{x^2-1}=3+\frac3{x^2-1}$$As $f$ is even, we may suppose $x\ge 0,\enspace x\ne q 1$.
Now the range of $x^2-1$ is $[-1,+\infty)$ and for $f(x)$, the value $x^2=1$ is excluded, hence the range of $\;\dfrac3{x^2-1}$ is $(-\infty,-3] $ for $x\in [0,1)$ and $(0,+\infty)$ for $x>1$.
Hence the range of $f(x)=3+\dfrac3{x^2-1}$ is $\;(-\infty,0]\cup (3,+\infty)$.
$\endgroup$ $\begingroup$Hint:
find for what values of $k$ the equation $k=\dfrac{3x^2}{x^2-1}$ has real solutions.
For $ x\ne \pm 1$ you have $ x^2= \dfrac{k}{k-3}$ so this fraction must be not negative: $\dfrac{k}{k-3}\ge 0$. The solution is the range.
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