NSW HSC Mathematics Extension 1

## The graph of the reciprocal, y = 1/f(x) The graph of $y = \dfrac{1}{f(x)}$ can be sketched from the graph of $y = f(x)$ without any calculation, by reading off a few features of $f$. | Feature of $y = f(x)$ | What happens to $y = \dfrac{1}{f(x)}$ | |---|---| | a zero, $f(a) = 0$ | a vertical asymptote at $x = a$ | | a vertical asymptote at $x = a$ | the graph approaches $0$ as $x \to a$ | | $f(x) \to \infty$ or $f(x) \to -\infty$ | $\dfrac{1}{f(x)} \to 0$ | | $f(x) \to 0$ (a horizontal asymptote at the $x$-axis) | $\dfrac{1}{f(x)}$ becomes unbounded | | $f(x) = 1$ or $f(x) = -1$ | the two graphs meet, because $\dfrac{1}{1} = 1$ and $\dfrac{1}{-1} = -1$ | | $f(x) > 0$ | $\dfrac{1}{f(x)} > 0$ | | $f(x) < 0$ | $\dfrac{1}{f(x)} < 0$ | The two graphs have the same sign everywhere, so the reciprocal never crosses the $x$-axis. The only points where the graphs meet are where $f(x)^2 = 1$, because $\dfrac{1}{f(x)} = f(x)$ rearranges to $f(x)^2 = 1$. Where $f$ is positive and increasing, its reciprocal is positive and decreasing, and where $f$ is negative and increasing, its reciprocal is also decreasing. A maximum of $f$ at a positive value becomes a minimum of the reciprocal, and a minimum of $f$ at a positive value becomes a maximum. ### Worked example Sketch $y = \dfrac{1}{x^2 - 4}$ from $f(x) = x^2 - 4$. The zeroes of $f$ are $x = \pm 2$, so the reciprocal has vertical asymptotes at $x = \pm 2$. As $x \to \pm\infty$, $f(x) \to \infty$, so $y \to 0$ from above. The function…