Section 5.1 Functions
Subsection 1. New vocabulary
Subsubsection Definitions
Write a definition or description for each term. You can find answers in Section 1.2 of your textbook.
- Function
- Input variable
- Output variable
- Function value
- Function notation
Subsubsection Exercise
Checkpoint 5.1.
Identify each term above, or give an example, for this situation: At time \(t\) seconds, the height of a basketball above the ground, \(h\text{,}\) in feet, is given by
\begin{equation*}
~~h=-16t^2+20t+5\text{.}
\end{equation*}
Answer
- \(h\) is a function of \(t\text{.}\)
- The input variable is \(t\text{.}\)
- The output variable is \(h\text{.}\)
- The function value for \(t=1\) is \(h=9\text{.}\)
- \(\displaystyle h=f(t)\)
Subsection 2. Solve non-linear equations
To solve simple non-linear equations, we "undo" the operation performed on the variable.
Subsubsection Examples
Example 5.2.
Solve the equation \(~~5 \sqrt{t} = 83\)
Solution
To "undo" a square root, we square both sides of the equation. First, we isolate the square root.
\begin{align*}
\dfrac{5 \sqrt{t}}{\alert{5}} \amp = \dfrac{83}{\alert{5}} \amp \amp \blert{\text{Divide both sides by 5.}}\\
(\sqrt{t})^2 \amp =(16.6)^2 \amp \amp \blert{\text{Square both sides.}}\\
t \amp =275.56
\end{align*}
Example 5.3.
Solve the equation \(~~\dfrac{15}{y}=45\)
Solution
If the variable is in the denominator of a fraction, we must first clear the fraction.
\begin{align*}
\alert{y}(\dfrac{15}{y}) \amp = 45 \cdot \alert{y} \amp \amp \blert{\text{Multiply both sides by} ~y.}\\
15 \amp = 45y \amp \amp \blert{\text{Divide both sides by 45.}}\\
y \amp =\dfrac{15}{45}=\dfrac{1}{3}
\end{align*}
Subsubsection Exercises
Checkpoint 5.4.
Solve the equation \(~~\dfrac{4.8}{w}=3\)
Answer
\(1.6\)
Checkpoint 5.5.
Solve the equation \(~~18=36\sqrt{q}\)
Answer
\(\dfrac{1}{4}\)