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Section 3.4 : The Definition of a Function
For problems 1 – 3 determine if the given relation is a function.
- \(\left\{ {\left( {2,4} \right),\left( {3, - 7} \right),\left( {6,10} \right)} \right\}\) Solution
- \(\left\{ {\left( { - 1,8} \right),\left( {4, - 7} \right),\left( { - 1,6} \right),\left( {0,0} \right)} \right\}\) Solution
- \(\left\{ {\left( {2,1} \right),\left( {9,10} \right),\left( { - 4,10} \right),\left( { - 8,1} \right)} \right\}\) Solution
For problems 4 – 6 determine if the given equation is a function.
- \(\displaystyle y = 14 - \frac{1}{3}x\) Solution
- \(y = \sqrt {3{x^2} + 1} \) Solution
- \({y^4} - {x^2} = 16\) Solution
- Given \(f\left( x \right) = 3 - 2{x^2}\) determine each of the following.
- \(f\left( 0 \right)\)
- \(f\left( 2 \right)\)
- \(f\left( { - 4} \right)\)
- \(f\left( {3t} \right)\)
- \(f\left( {x + 2} \right)\)
- Given \(\displaystyle g\left( w \right) = \frac{4}{{w + 1}}\) determine each of the following.
- \(g\left( { - 6} \right)\)
- \(g\left( { - 2} \right)\)
- \(g\left( 0 \right)\)
- \(g\left( {t - 1} \right)\)
- \(g\left( {4w + 3} \right)\)
- Given \(h\left( t \right) = {t^2} + 6\) determine each of the following.
- \(h\left( 0 \right)\)
- \(h\left( { - 2} \right)\)
- \(h\left( 2 \right)\)
- \(h\left( {\sqrt x } \right)\)
- \(h\left( {3 - t} \right)\)
- Given \(h\left( z \right) = \left\{ {\begin{array}{*{20}{l}}{3z}&{{\rm{if }}z < 2}\\{1 + {z^2}}&{{\rm{if }}z \ge 2}\end{array}} \right.\) determine each of the following.
- \(h\left( 0 \right)\)
- \(h\left( 2 \right)\)
- \(h\left( 7 \right)\)
- 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.
- \(f\left( { - 4} \right)\)
- \(f\left( 2 \right)\)
- \(f\left( 6 \right)\)
- \(f\left( 9 \right)\)
- \(f\left( {12} \right)\)
For problems 12 & 13 compute the difference quotient for the given function. The difference quotient for the function \(f\left( x \right)\) is defined to be,
\[\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]For problems 14 – 18 determine the domain of the function.
- \(A\left( x \right) = 6x + 14\) Solution
- \(\displaystyle f\left( x \right) = \frac{1}{{{x^2} - 25}}\) Solution
- \(\displaystyle g\left( t \right) = \frac{{8t - 24}}{{{t^2} - 7t - 18}}\) Solution
- \(g\left( w \right) = \sqrt {9w + 7} \) Solution
- \(\displaystyle f\left( x \right) = \frac{1}{{\sqrt {{x^2} - 8x + 15} }}\) Solution