Modulus of functions and reciprocal graphsAQA A-Level Further Maths: Revision notes

The modulus of a number is its distance from zero: $|x|=x$ if $x\ge0$ and $|x|=-x$ if $x<0$, so $|x|$ is never negative. For a function, $|\mathrm{f}(x)|=\mathrm{f}(x)$ where $\mathrm{f}(x)\ge0$ and $-\mathrm{f}(x)$ where $\mathrm{f}(x)<0$. To solve $|\mathrm{f}(x)|=k$ with $k>0$, solve both $\mathrm{f}(x)=k$ and $\mathrm{f}(x)=-k$. For $|2x-1|=5$: $x=3$ or $x=-2$. To solve $|\mathrm{f}(x)|=\mathrm{g}(x)$, solve $\mathrm{f}(x)=\mathrm{g}(x)$ and $\mathrm{f}(x)=-\mathrm{g}(x)$, then check each answer in the original equation, because $|\ |$ cannot equal a negative value. For $|2x-1|=x+4$: $x=5$ and $x=-1$ both check.

For $k>0$:

• $|\mathrm{f}(x)|<k\iff-k<\mathrm{f}(x)<k$ (one interval).
• $|\mathrm{f}(x)|>k\iff\mathrm{f}(x)<-k$ or $\mathrm{f}(x)>k$ (two intervals). Example: $|2x-1|<3\Rightarrow-3<2x-1<3\Rightarrow-1<x<2$. When comparing two moduli, square both sides: $|\mathrm{f}(x)|<|\mathrm{g}(x)|\iff\mathrm{f}(x)^2<\mathrm{g}(x)^2$, then factorise the difference of two squares. Example: $|x^2-4x|<5$ means $-5<x^2-4x<5$. The right side gives $-1<x<5$; the left side, $x^2-4x+5>0$, is always true since $(x-2)^2+1>0$. The solution is $-1<x<5$.

To sketch $y=|\mathrm{f}(x)|$, draw $y=\mathrm{f}(x)$, then reflect in the $x$-axis every part that lies below it. Parts already on or above the axis stay. The graph never goes below the $x$-axis, and where $\mathrm{f}$ crosses the axis the new graph has a sharp corner. For $\mathrm{f}(x)=x^2-6x+5$ the minimum $(3,-4)$ becomes a local maximum $(3,4)$ of $|\mathrm{f}|$. The equation $|\mathrm{f}(x)|=k$ then has four solutions for $0<k<4$, three for $k=4$ and two for $k>4$. Algebraically, intersections of $y=|\mathrm{f}(x)|$ with a line come from both $\mathrm{f}(x)=k$ and $\mathrm{f}(x)=-k$.

Features of $y=\frac1{\mathrm{f}(x)}$ come from $y=\mathrm{f}(x)$:

• Vertical asymptotes wherever $\mathrm{f}(x)=0$.
• The sign is unchanged: where $\mathrm{f}$ is positive, so is the reciprocal.
• Points where $\mathrm{f}(x)=\pm1$ stay on the graph (reciprocal of $1$ is $1$).
• Where $\mathrm{f}$ is large, the reciprocal is close to $0$, so $y=0$ is a horizontal asymptote when $\mathrm{f}(x)\to\pm\infty$.
• A minimum of $\mathrm{f}$ with $\mathrm{f}>0$ becomes a maximum of the reciprocal, and vice versa. A minimum of $-4$ gives a maximum of $-\frac14$.
• The $y$-intercept is $\frac{1}{\mathrm{f}(0)}$, and the reciprocal never crosses the $x$-axis.

Example: $\mathrm{f}(x)=x^2+2x-3=(x-1)(x+3)$. The graph of $y=\frac1{\mathrm{f}(x)}$ has vertical asymptotes $x=1$ and $x=-3$, horizontal asymptote $y=0$, $y$-intercept $-\frac13$, and a local maximum at $\left(-1,-\frac14\right)$ because $\mathrm{f}$ has minimum $-4$ at $x=-1$. Its range is $y\le-\frac14$ or $y>0$: between the asymptotes $\mathrm{f}\in[-4,0)$, giving $\frac1{\mathrm{f}}\le-\frac14$; outside them $\mathrm{f}>0$ gives $\frac1{\mathrm{f}}>0$. When asked for a range, work from the range of $\mathrm{f}$, remembering that $0$ is never taken by $\frac1{\mathrm{f}}$.