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Section 1-1 : Review : Functions

2. Perform the indicated function evaluations for \(\displaystyle g\left( t \right) = \frac{t}{{2t + 6}} \).

  1. \(g\left( 0 \right) \)
  2. \(g\left( { - 3} \right)\)
  3. \(g\left( {10} \right) \)
  1. \(g\left( {{x^2}} \right) \)
  2. \(g\left( {t + h} \right)\)
  3. \(g\left( {{t^2} - 3t + 1} \right) \)

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a \(g\left( 0 \right) \) Show Solution
\[g\left( 0 \right) = \frac{0}{{2\left( 0 \right) + 6}} = \frac{0}{6} = 0\]

b \(g\left( { - 3} \right)\) Show Solution
\[\require{cancel}g\left( -3 \right)=\frac{-3}{2\left( -3 \right)+6}=\frac{-3}{0}\,\,\,\Huge \times \]

The minute we see the division by zero we know that \(g\left( { - 3} \right)\) does not exist.


c \(g\left( {10} \right) \) Show Solution
\[g\left( {10} \right) = \frac{{10}}{{2\left( {10} \right) + 6}} = \frac{{10}}{{26}} = \frac{5}{{13}}\]

Hint : Don’t let the fact that there are now variables here instead of numbers get you confused. This works exactly the same way as the first three it will just have a little more algebra involved.
d \(g\left( {{x^2}} \right) \) Show Solution
\[g\left( {{x^2}} \right) = \frac{{{x^2}}}{{2{x^2} + 6}}\]

Hint : Don’t let the fact that there are now variables here instead of numbers get you confused. Also, don’t get excited about the fact that there is both a \(t\) and an \(h\) here. This works exactly the same way as the first three it will just have a little more algebra involved.
e \(g\left( {t + h} \right)\) Show Solution
\[g\left( {t + h} \right) = \frac{{t + h}}{{2\left( {t + h} \right) + 6}} = \frac{{t + h}}{{2t + 2h + 6}}\]

Hint : Don’t let the fact that there are now variables here instead of numbers get you confused. This works exactly the same way as the first three it will just have a little more algebra involved.
f \(g\left( {{t^2} - 3t + 1} \right) \) Show Solution
\[g\left( {{t^2} - 3t + 1} \right) = \frac{{{t^2} - 3t + 1}}{{2\left( {{t^2} - 3t + 1} \right) + 6}} = \frac{{{t^2} - 3t + 1}}{{2{t^2} - 6t + 8}}\]