I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 3.4 : The Definition of a Function
8. Given \(\displaystyle g\left( w \right) = \frac{4}{{w + 1}}\) determine each of the following.
- \(g\left( { - 6} \right)\)
- \(g\left( { - 2} \right)\)
- \(g\left( 0 \right)\)
- \(g\left( {t - 1} \right)\)
- \(g\left( {4w + 3} \right)\)
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a \(g\left( { - 6} \right)\) Show SolutionRemember 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,
\[g\left( {4w + 3} \right) = \frac{4}{{\left( {4w + 3} \right) + 1}} = \frac{4}{{4w + 4}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{w + 1}}}}\]