Question

Let f(x) = x/abs((x^2)-9)
Find the domain of f.
Write an equation for each vertical/horizontal asymptote of graph of f.
Is f odd, even, or neither? Justify your answer.
find all values of x for which f is discontinuous and classify each discontinuity as removable or nonremovable.

Answer 1

first f is odd since f(-x)=-f(x)
f is discontinous if x^2=9 i.e., x=3or x=-3
I think it is a non removable discontinuity
domain of f is R-{-3,3}

Answer 2

x=3 and x=-3 are asymptodes
i am not sure but i think y=0 will be an asymptode too.

Answer 3

\[\lim_{x \rightarrow \infty}\frac{x}{\sqrt{(x^2-9)^2}}=\lim_{x \rightarrow \infty}\frac{x}{\sqrt{x^4-18x^2+81}}\]
\[=\lim_{x \rightarrow \infty}\frac{\frac{1}{\sqrt{x^4}}}{\frac{1}{\sqrt{x^4}}} \cdot \frac{x}{\sqrt{x^4-18x^2+81}}\]
\[\sqrt{x^4}|=|x^2|=x^2 \text{ for all } x \text{ since } x^2 \ge 0\]
\[\lim_{x \rightarrow \infty} \frac{\frac{x}{x^2}}{\sqrt{\frac{x^4-18x^2+81}{x^4}}}\]
\[\frac{0}{\sqrt{1-0+0}}=0\]
y=0 is the horizontal asymptote