I have the following function:
$y=f(x)=\frac{1}{1-x^{2}}$ for $x \in M=(-1,1)$
How would I find the bounds? I tried setting various numbers from the interval as $X$ and I was able to see that there is not upper bound, as the function can rise to positive infinity (1/0,00001..), however my assumption was that the lower bound is 1, which is incorrect.
Can you tell me how to find the upper and lower bounds in general (without guessing)? Thanks
Computing the derivative yields $$ f'(x) = \frac{2x}{1-x^2} $$ which has a unique solution at $x=0$ where $f(0) = 1$. See that $f(x) > 1$ for all $x \in M \setminus \{0\}$. Thus you have that $1$ is a lower bound. To see that it is not bounded from above compute the limit $$ \lim_{x \to 1^-}f(x)=\infty $$ and conclude from this that there is no $c\in\mathbb{R}$ such that $f(x) \le c$ for all $x \in M$.