2.16 Modulus and further transformed graphsIB Maths: Analysis and Approaches HL: Revision notes

$y=|f(x)|$: keep the parts of $y=f(x)$ on or above the $x$-axis and reflect the parts below the $x$-axis in the $x$-axis. Zeros stay where they are; a minimum below the axis becomes a maximum, e.g. $(1,-9)\to(1,9)$. The range contains no negative values.

$y=f(|x|)$: keep the part of $y=f(x)$ for $x\ge0$ and reflect it in the $y$-axis; the part for $x<0$ is discarded. The result is always an even function. For $f(x)=x^{2}-2x-8$, the zero at $x=4$ is kept and copied to $x=-4$; the zero at $x=-2$ disappears.

Key features follow from those of $f$:

• Zeros of $f$ become vertical asymptotes of $\frac{1}{f}$.
• Vertical asymptotes of $f$ become zeros of $\frac{1}{f}$ (the graph approaches $0$).
• A local maximum $(a,k)$ with $k\neq0$ becomes a local minimum $\left(a,\frac{1}{k}\right)$, and vice versa.
• Where $f\to\pm\infty$, $\frac{1}{f}\to0$, so a horizontal asymptote $y=0$ appears.
• The sign is unchanged: $\frac{1}{f}$ is positive where $f$ is positive.
• Points where $f=\pm1$ are invariant.

$y=f(ax+b)$: to find where a feature at $x=c$ moves, solve $ax+b=c$, so $x=\frac{c-b}{a}$. The $y$-values do not change. For example, a minimum of $h$ at $(2,-3)$ moves to $(1.5,-3)$ on $y=h(2x-1)$, because $2x-1=2$. This is a horizontal translation by $-b$ followed by a horizontal stretch with scale factor $\frac{1}{a}$.

$y=[f(x)]^{2}$: every $y$-value is squared, so the graph is never below the $x$-axis. Zeros of $f$ remain zeros (the graph touches the axis there), points with $y=0$ or $y=1$ are invariant, values with $|f|<1$ get smaller and values with $|f|>1$ get larger. If $-1\le f\le3$ then $0\le[f]^{2}\le9$.

To solve $|A|=|B|$ (or $|A|=B$):

• Case method: $A=B$ or $A=-B$.
• Squaring method: $A^{2}=B^{2}$, valid because both sides are non-negative.

Example: $|2x-5|=|x+1|$ gives $x=6$ or $x=\frac{4}{3}$.

For $|A|=B$ where $B$ could be negative, always check each solution. For $|x-1|=2x$: the case $x-1=2x$ gives $x=-1$, which fails because $2x=-2<0$; the case $x-1=-2x$ gives $x=\frac{1}{3}$, which works. So $x=\frac{1}{3}$ only.

For $k>0$: $|A|<k\iff-k<A<k$, and $|A|>k\iff A<-k$ or $A>k$.

For $|A|>|B|$, square both sides (both are non-negative): $A^{2}>B^{2}$, then solve the resulting quadratic inequality. Example: $|2x-5|>|x+1|\iff(3x-4)(x-6)>0\iff x<\frac{4}{3}$ or $x>6$.

For harder inequalities, such as $|3x\arccos x|>1$, use technology: graph both sides and find the intersections, remembering any domain restriction (here $-1\le x\le1$).