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!