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 6.1 : Exponential Functions
2. Given the function \(f\left( x \right) = {\left( {\displaystyle \frac{1}{5}} \right)^x}\) evaluate each of the following.
- \(f\left( { - 3} \right)\)
- \(f\left( { - 1} \right)\)
- \(f\left( 0 \right)\)
- \(f\left( 2 \right)\)
- \(f\left( 3 \right)\)
Show All Solutions Hide All Solutions
a \(f\left( { - 3} \right)\) Show SolutionAll we need to do here is plug in the \(x\) and do any quick arithmetic we need to do.
\[f\left( { - 3} \right) = {\left( {\frac{1}{5}} \right)^{ - 3}} = {\left( {\frac{5}{1}} \right)^3} = \frac{{{5^3}}}{{{1^3}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{125}}\]b \(f\left( { - 1} \right)\) Show Solution
All we need to do here is plug in the \(x\) and do any quick arithmetic we need to do.
\[f\left( - \right) = {\left( {\frac{1}{5}} \right)^{ - \,1}} = {\left( {\frac{5}{1}} \right)^1} = \require{bbox} \bbox[2pt,border:1px solid black]{5}\]c \(f\left( 0 \right)\) Show Solution
All we need to do here is plug in the \(x\) and do any quick arithmetic we need to do.
\[f\left( 0 \right) = {\left( {\frac{1}{5}} \right)^0} = \require{bbox} \bbox[2pt,border:1px solid black]{1}\]d \(f\left( 2 \right)\) Show Solution
All we need to do here is plug in the \(x\) and do any quick arithmetic we need to do.
\[f\left( 2 \right) = {\left( {\frac{1}{5}} \right)^2} = \frac{{{1^2}}}{{{5^2}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{25}}}}\]e \(f\left( 3 \right)\) Show Solution
All we need to do here is plug in the \(x\) and do any quick arithmetic we need to do.
\[f\left( 3 \right) = {\left( {\frac{1}{5}} \right)^3} = \frac{{{1^3}}}{{{5^3}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{125}}}}\]