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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
$g\left( {{t^2} - 3t + 1} \right) = \frac{{{t^2} - 3t + 1}}{{2\left( {{t^2} - 3t + 1} \right) + 6}} = \frac{{{t^2} - 3t + 1}}{{2{t^2} - 6t + 8}}$