Paul's Online Notes
Home / Algebra / Graphing and Functions / The Definition of a Function
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 3.4 : The Definition of a Function

8. Given $$\displaystyle g\left( w \right) = \frac{4}{{w + 1}}$$ determine each of the following.

1. $$g\left( { - 6} \right)$$
2. $$g\left( { - 2} \right)$$
3. $$g\left( 0 \right)$$
4. $$g\left( {t - 1} \right)$$
5. $$g\left( {4w + 3} \right)$$

Show All Solutions Hide All Solutions

a $$g\left( { - 6} \right)$$ Show Solution

Remember that for function evaluation all we need to do is replace all the $$w$$’s (only one in this case) with the number that is in the parenthesis. Doing this for this part gives,

$g\left( { - 6} \right) = \frac{4}{{ - 6 + 1}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{4}{5}}}$

b $$g\left( { - 2} \right)$$ Show Solution

Remember that for function evaluation all we need to do is replace all the $$w$$’s (only one in this case) with the number that is in the parenthesis. Doing this for this part gives,

$g\left( { - 2} \right) = \frac{4}{{ - 2 + 1}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 4}}$

c $$g\left( 0 \right)$$ Show Solution

Remember that for function evaluation all we need to do is replace all the $$w$$’s (only one in this case) with the number that is in the parenthesis. Doing this for this part gives,

$g\left( 0 \right) = \frac{4}{{0 + 1}} = \require{bbox} \bbox[2pt,border:1px solid black]{4}$

d $$g\left( {t - 1} \right)$$ Show Solution

Remember that for function evaluation all we need to do is replace all the $$w$$’s (only one in this case) with whatever is in the parenthesis. It doesn’t matter if the stuff in the parenthesis is not a number the evaluation works exactly the same as if it was a number.

Doing the evaluation for this part gives,

$g\left( {t - 1} \right) = \frac{4}{{\left( {t - 1} \right) + 1}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{4}{t}}}$

e $$g\left( {4w + 3} \right)$$ Show Solution

Remember that for function evaluation all we need to do is replace all the $$w$$’s (only one in this case) with whatever is in the parenthesis. It doesn’t matter if the stuff in the parenthesis is not a number the evaluation works exactly the same as if it was a number. Do not get excited about the fact that the stuff in the parenthesis also involves an $$w$$. Again, it does not matter, the evaluation still works the same way.

Doing the evaluation for this part gives,

$g\left( {4w + 3} \right) = \frac{4}{{\left( {4w + 3} \right) + 1}} = \frac{4}{{4w + 4}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{w + 1}}}}$