$2x + 1 \le 2$ $ \ \ \ \ -1 = -1$ $2x\ \ \ \ \ \le 1$ $\frac 12 (2x) \le \frac 12 (1)$ $\ \ \ \ \ x \le \frac 12$ |

This conforms with the symbolic result, $x \le \frac {C- B} A$ where $A=2, B = 1$, and $C = 2$. So the solution is $ x \le \frac {2- 1} 2 = \frac {1} 2.$

The problem can be restated now: To find a value for $x$ where $f(x) \le 2$.

This problem and its solution can be visualized with mapping diagrams for the linear functions $f$ treated as a composition of $m$ followed by $s$.

On the GeoGebra mapping diagram for the composition, find $2$ on the final target axis.

Undo $s$ by subtracting $1$ from $2$. This corresponds to going back on the diagram from $2$ on the final axis to $2-1=1$ on the middle axis, reversing the arrow that would go from $1$ to $2$ visualizing $s$.

Thus if $m(x) = 1$ then $f(x) = s(m(x))=s(1)=2$

Undo $m$ by multiplying $1$ by $\frac 12$. This corresponds to going back on the diagram from $1$ on the middle axis to $\frac 12$ on the source axis, reversing the arrow that would go from $\frac 12$ to $1$ visualizing $m$.

Thus when $x\le\frac 12$, $f(x) \le s(m(\frac 12))=s(1)=2$

On the GeoGebra sketch you can change the values of $m=A$ and $b=B$ (with the sliders) and $C$ (currently $2$ on the target axis) to see other linear inequalities $Ax + B \le C$ are solved visually with the mapping diagram.

Undo $s$ by subtracting $1$ from $2$. This corresponds to going back on the diagram from $2$ on the final axis to $2-1=1$ on the middle axis, reversing the arrow that would go from $1$ to $2$ visualizing $s$.

Thus if $m(x) = 1$ then $f(x) = s(m(x))=s(1)=2$

Undo $m$ by multiplying $1$ by $\frac 12$. This corresponds to going back on the diagram from $1$ on the middle axis to $\frac 12$ on the source axis, reversing the arrow that would go from $\frac 12$ to $1$ visualizing $m$.

Thus when $x\le\frac 12$, $f(x) \le s(m(\frac 12))=s(1)=2$

On the GeoGebra sketch you can change the values of $m=A$ and $b=B$ (with the sliders) and $C$ (currently $2$ on the target axis) to see other linear inequalities $Ax + B \le C$ are solved visually with the mapping diagram.