Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.3 Exercises - Page 345: 86

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Although the function has a $finite$ negative slope $almost$* everywhere (which normally means that it is strictly increasing and is hence one-one), it is discontinuous** as can be seen in the graph of $y=\frac{x}{x^2-4}$. So, it is not one-to-one. For example, $f(-1) = f(4) =\frac{1}{3}$ *At the points of discontinuity, the slope is not finite **It is discontinuous when $x^2-4=0$, i.e at $x=-2$ and at $x=+2$