output \(\rightarrow \) input

We are tracing individual domain numbers through function chains as our first look into function composition. When each function uses only one formula, then following the trail requires no decisions. Now we turn to piecewise defined functions where we need to pick the appropriate formula.

Graphically

The functions \(V\) and \(T\) are defined graphically via their graphs below.

Piecewise Defined Formulas

Use the graphs above to approimate the following expressions.

\[ \begin{array}{l|l} V(6) = \answer [tolerance=0.1]{2} & T(V(6)) = \answer [tolerance=0.1]{1} \\ T(5) = \answer [tolerance=0.1]{-2} & V(T(5)) = \answer [tolerance=0.1]{1} \\ V(7) = \answer [tolerance=0.1]{4} & V(V(7)) = \answer [tolerance=0.1]{-7} \\ T(1) = \answer [tolerance=0.1]{2} & T(T(1)) = \answer [tolerance=0.1]{1} \\ T(-5) = \answer [tolerance=0.1]{4} & T(T(-5)) = \answer [tolerance=0.1]{-1} \\ \end{array} \]

Algebraically

The functions \(f\) and \(g\) are defined via the formulas below.

\[ f(y) = \begin{cases} -2y-3 & [-6, -2] \\ -y-3 & (-2, 4] \\ 2y - 10 & (4,8) \end{cases} \]
\[ g(w) = \begin{cases} -2w-6 & -7 < w < 1 \\ -w+3 & 1 \leq w < 7 \end{cases} \]
Piecewise Defined Formulas

Use the formulas above to evaluate the following expressions.

\[ \begin{array}{l|l} f(5) = \answer {0} & g(f(5)) = \answer {-6} \\ f(7) = \answer {4} & g(f(7)) = \answer {-1} \\ g(-1) = \answer {-4} & f(g(-1)) = \answer {5} \\ g(-3) = \answer {0} & f(g(-3)) = \answer {-3} \\ g(-6) = \answer {6} & f(g(-6)) = \answer {2} \\ f(5) = \answer {0} & f(f(5)) = \answer {-3} \\ f(0) = \answer {-3} & f(f(0)) = \answer {3} \\ f(4) = \answer {-7} & f(f(4)) = \answer {DNE} \\ g(6) = \answer {-3} & g(g(6)) = \answer {0} \\ g(-6) = \answer {6} & g(g(-6)) = \answer {-3} \\ g(-1) = \answer {-4} & g(g(-1)) = \answer {2} \\ \end{array} \]

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more examples can be found by following this link
More Examples of Pointwise Composition

2025-01-07 00:29:04