MHB Give an example of a function f(x)

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A function example that achieves the range \( f([-1, 1]) = (-\infty, \infty) \) is \( f(x) = \frac{x}{(x-1)(x+1)} \). Another proposed function is \( f(x) = \tan\left(\frac{x}{\pi/2}\right) \), which maps \( (-1, 1) \) to \( (-\infty, \infty) \). However, for the closed interval \( [-1, 1] \), additional consideration is needed, such as defining \( f(-1) = f(1) = 0 \). These functions demonstrate the ability to achieve the desired output range. The discussion highlights the importance of function behavior over specified intervals.
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Give an example of a function $\displaystyle f(x)$ for which $\displaystyle f([-1,\ 1])=(-\infty,\ \infty)$.

My thoughts: $\displaystyle f(x)=\frac{x}{(x-1)(x+1)}$ is a function for which $f((-1,\ 1))=(-\infty,\ \infty)$.
 
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Alexmahone said:
Give an example of a function $\displaystyle f(x)$ for which $\displaystyle f([-1,\ 1])=(-\infty,\ \infty)$.

My thoughts: $\displaystyle f(x)=\frac{x}{(x-1)(x+1)}$ is a function for which $f((-1,\ 1))=(-\infty,\ \infty)$.

\(\displaystyle f(x)=\tan \left( \frac{x}{\pi/2} \right), \ \ x\in (-1,1) \)

will map \( (-1,1) \) to \( (-\infty, \infty) \) but \( [-1,1] \) might need a bit more thought, though you can just define \( f(-1)=f(1)=0 \)

CB
 
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