What an inequality solution really is

A solution set contains the input values that make a mathematical statement true. Here, that statement will usually be

On a graph, the question becomes: for which $x$-values does $y=g(x)$ lie on or above $y=f(x)$?

The word on matters. With $\geq$ or $\leq$, include the intersection points. With $>$ or $<$, leave them out.

The graphical method

To solve $g(x) \geq f(x)$ graphically, sketch or display both curves on the same axes. Locate the intersection points by solving $g(x)=f(x)$ where possible, or use technology. Next, read off the intervals where the graph of $g$ lies above the graph of $f$.

Some functions don’t give a neat exact solution, so technology is the sensible tool rather than a shortcut around understanding. You still have to decide which curve is higher between the intersection points. The domain also matters. For example, when a logarithm appears, don’t include any values of $x$ for which it isn’t defined.

Choose the graphing window carefully. A poor window may hide an intersection or make two curves appear to touch when they don’t. Representation matters here: algebra shows the exact structure, while the graph shows where the comparison takes place.

The analytic method for simple polynomials

For simple polynomial inequalities up to degree $3$, a clean analytic approach is to move everything to one side. Define

Then

Now solve $h(x)=0$. Its roots are boundary points that divide the real line into intervals. You can test one value in each interval, or use the shape and sign changes of the factorised expression.

For example, if

then the boundary points are $x=-4$, $x=-1$ and $x=2$. At simple roots, the sign of $h(x)$ alternates, so a sign chart shows where $h(x)\geq0$. If a factor is repeated, the graph touches the $x$-axis there and the sign does not change. Students forget this more often than they should.

Inequality algebra: be careful with signs

Adding or subtracting the same expression on both sides keeps the inequality direction unchanged. The same applies when multiplying or dividing by a positive quantity. Multiply or divide by a negative quantity, however, and the direction reverses.

The rule itself isn’t the difficult part. The danger is forgetting to check whether the quantity you’re dividing by is positive. An expression involving $x$ may change sign across its domain. If that happens, don’t divide casually; move everything to one side and work with intervals.

For instance, since $\ln\left(\frac{1}{2}\right)<0$, solving

requires you to reverse the inequality direction when dividing by $\ln\left(\frac{1}{2}\right)$.

A note on mathematical value

The TOK question linked to this statement is worth considering. Some mathematical cultures and classrooms value exact algebraic solutions, while others favour visual or technological insight. IB questions often reward both approaches in practice. The best solution should be justified and efficient, and it should suit the function in front of you.

The two places a transformation can act

A graph transformation changes the equation of a function in a way that produces a predictable change in its graph. A change outside $f$ acts vertically; a change inside the input of $f$ acts horizontally. Keeping that distinction clear prevents muddled sketches.

A modulus gives a function value’s distance from zero on the number line, so it can never be negative. For an expression

$y=|f(x)|$: fold negative outputs upwards

In $y=|f(x)|$, the modulus sits outside the function. The input $x$ doesn’t change. Instead, every negative output becomes positive. Any part of $y=f(x)$ below the $x$-axis is reflected in the $x$-axis, while points on or above the $x$-axis stay in place.

Zeros of $f$ remain zeros. After reflection, a local maximum below the $x$-axis becomes a local minimum, while a local minimum below the $x$-axis becomes a local maximum. Across its domain, the transformed graph has $y\geq0$.

$y=f(|x|)$: copy the right-hand side to the left

Here, in $y=f(|x|)$, the modulus is inside the function. Since the function receives $|x|$ as its input, only the original graph’s behaviour for non-negative inputs is used. The section of $y=f(x)$ for $x\geq0$ stays where it is, then a reflection of that section in the $y$-axis creates the graph for $x<0$.

Wherever it is defined, the result is an even function: replacing $x$ by $-x$ leaves $|x|$ unchanged. Don’t use the original left-hand side of $f$ because it has been overwritten.

$y=\frac{1}{f(x)}$: reciprocals change size and create asymptotes

For $y=\frac{1}{f(x)}$, each output of $f$ is replaced by its reciprocal. Any $x$-value for which $f(x)=0$ must be excluded from the domain, since division by zero is undefined. These zeros of $f$ become vertical asymptotes on the reciprocal graph.

The sign stays the same. Positive values of $f(x)$ produce positive reciprocals, while negative values produce negative reciprocals. Large values of $|f(x)|$ turn into values close to zero. By contrast, when $f(x)$ is close to zero, its reciprocal becomes very large in magnitude. At any point where $f(x)=1$ or $f(x)=-1$, the reciprocal and original graphs pass through the same point.

If $f$ has a horizontal asymptote $y=k$, where $k$ is a non-zero constant, the reciprocal graph has horizontal asymptote $y=\frac{1}{k}$. If $f(x)\to\pm\infty$, then $\frac{1}{f(x)}\to0$.

$y=f(ax+b)$: a horizontal inside transformation

Solving for the new input gives

This rule changes the $x$-coordinates but leaves the $y$-coordinates unchanged. When $a>0$, the graph undergoes a horizontal scale factor of $\frac{1}{a}$ together with a horizontal shift. When $a<0$, it is also reflected in the $y$-axis. Dynamic graphing works especially well in calculator work here: move $a$ and $b$, then watch the graph compress, stretch, shift and reflect.

$y=[f(x)]^2$: square the outputs

For $y=[f(x)]^2$, the whole output of $f$ is squared, making this a vertical transformation. The new graph has $y\geq0$. Zeros stay as zeros, and every point where $f(x)=1$ or $f(x)=-1$ moves to height $1$.

When $0<|f(x)|<1$, squaring reduces the output and brings it closer to the $x$-axis. When $|f(x)|>1$, the output becomes larger and moves farther from the $x$-axis. Negative sections of the original graph become positive, but this isn’t simply a reflection unless the original value is exactly matched in magnitude. Squaring changes both the sign and the scale.

Analytic and visual thinking

The guide’s international-mindedness link contrasts a strongly analytic tradition with a strongly visual one. These transformations show why both approaches help. A formula provides the exact rule, while a graph reveals its spatial effect. You need to be comfortable using both languages.

The basic idea

A modulus equation is an equation with at least one modulus expression, such as $|x-3|$ or $|2x+1|$. A modulus inequality is an inequality that contains at least one modulus expression.

The modulus symbol changes the rule according to the sign of the expression inside it.

For example,

so

The condition changes as well. Don’t simply remove the modulus sign; remove it only on the stated region.

Solving modulus equations analytically

A reliable analytic approach is to split the number line wherever an expression inside a modulus sign equals zero. On the modulus graphs, these values are the bounce points.

Take an equation involving $|x+3|$ and $\left|\frac{1}{2}x-1\right|$. Its critical values are $x=-3$ and $x=2$, creating three regions: $x<-3$, $-3\leq x<2$ and $x\geq2$. In each region, replace every modulus expression with the appropriate linear expression and solve. Then check whether the answer actually lies in the region you used.

That check is essential. A value obtained from the wrong region doesn’t solve the original equation.

Useful exact modulus facts

When $k<0$, the equation $|A|=k$ has no solution because a modulus cannot be negative. When $k\geq0$,

For inequalities, use the standard forms

and

The non-strict versions work in the same way: replace $<$ with $\leq$ or $>$ with $\geq$. These facts are quick to apply. However, the piecewise method is safer if there is more than one modulus expression or the other side isn’t a constant.

Solving modulus inequalities graphically

For a modulus inequality, begin with the corresponding equality and solve it to find the boundary points. Next, use the graph to identify where the inequality holds. For example, suppose the question is

Graph $y=|x-1|$ and $y=\frac{1}{2}x+2$. The solution consists of the $x$-values for which the V-shaped modulus graph lies on or below the line.

Technology is particularly useful when there is no tidy analytic route. The guide’s example $|3x\arccos(x)|>1$ shows this well: the domain is $-1\leq x\leq1$, and a graphing package or calculator can locate both the boundary points and the intervals where the inequality holds.

What to watch when working by hand

Write down the regions before removing any modulus signs. For inequalities, multiplying or dividing by a negative expression reverses the inequality sign. If its sign is unclear, don’t divide by it. Instead, split into cases, or move everything to one side and use a graph or sign chart.

Strong hand solutions often use both approaches: piecewise algebra gives exact results, while a quick sketch checks the number of solutions and the intervals.