Algebra - The Definition of a Function

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Section 3.4 : The Definition of a Function

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11. Given \(f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}6&{{\rm{if }}x \ge 9}\\{x + 9}&{{\rm{if }}2 < x < 9}\\{{x^2}}&{{\rm{if }}x \le 2}\end{array}} \right.\) determine each of the following.

\(f\left( { - 4} \right)\)
\(f\left( 2 \right)\)
\(f\left( 6 \right)\)
\(f\left( 9 \right)\)
\(f\left( {12} \right)\)

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a \(f\left( { - 4} \right)\) Show Solution

Remember that for piecewise functions we use the equation for which the number in the parenthesis meets the condition.

For this problem we can see that \( - 4 \le 2\) and so we use bottom equation to do the evaluation.

\[f\left( { - 4} \right) = {\left( { - 4} \right)^2} = \require{bbox} \bbox[2pt,border:1px solid black]{{16}}\]

b \(f\left( 2 \right)\) Show Solution

Remember that for piecewise functions we use the equation for which the number in the parenthesis meets the condition.

For this problem we can see that \(2 \le 2\) and so we use bottom equation to do the evaluation.

\[f\left( 2 \right) = {\left( 2 \right)^2} = \require{bbox} \bbox[2pt,border:1px solid black]{4}\]

c \(f\left( 6 \right)\) Show Solution

Remember that for piecewise functions we use the equation for which the number in the parenthesis meets the condition.

For this problem we can see that \(2 < 6 < 9\) and so we use middle equation to do the evaluation.

\[f\left( 6 \right) = 6 + 9 = \require{bbox} \bbox[2pt,border:1px solid black]{{15}}\]

d \(f\left( 9 \right)\) Show Solution

Remember that for piecewise functions we use the equation for which the number in the parenthesis meets the condition.

For this problem we can see that \(9 \ge 9\) and so we use top equation to do the evaluation.

\[f\left( 9 \right) = \require{bbox} \bbox[2pt,border:1px solid black]{6}\]

Do not get excited about the fact that there are no \(x\)’s in the equation we are evaluating!

e \(f\left( {12} \right)\) Show Solution

Remember that for piecewise functions we use the equation for which the number in the parenthesis meets the condition.

For this problem we can see that \(12 \ge 9\) and so we use top equation to do the evaluation.

\[f\left( {12} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{6}\]

Do not get excited about the fact that there are no \(x\)’s in the equation we are evaluating!