9.3: Evaluating Functions
- Page ID
- 173881
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Find the Value of a Function
It is very convenient to name a function and most often we name it \(f\), \(g\), \(h\), \(F\), \(G\), or \(H\). In any function, for each \(x\)-value from the domain we get a corresponding \(y\)-value in the range. For the function \(f\), we write this range value \(y\) as \(f(x)\). This is called function notation and is read \(f\) of \(x\) or the value of \(f\) at \(x\). In this case the parentheses does not indicate multiplication.
For the notation \(y=f(x)\), \(f\) is the name of the function, \(x\) is the domain value, also known as the independent variable, and \(f(x)\) is the range value of the function \(f\) at \(x\). Since \(y = f(x)\), we can also say that the range value of the function \(f\) at \(x\) is \(y\). In this case, we say that \(y\) is the dependent variable (as its value depends on \(x\)). We read \(f(x)\) as "\(f\) of \(x\)" or the value of \(f\) at \(x\).
This process of finding the value of \(f(x)\) for a given value of \(x\) is called evaluating the function.
For the function \(f(x)=2x^2+3x−1\), evaluate the function.
- \(f(3)\)
- \(f(−2)\)
- \(f(a)\)
- Solutions
-
- \[ \begin{array}{rcl}
f(x) = 2x^2 + 3x - 1 & \implies & f(3) = 2(3)^2 + 3(3) - 1 \\[6pt]
& \implies & f(3) = 2(9) + 3(3) - 1 \\[6pt]
& \implies & f(3) = 18 + 9 - 1 \\[6pt]
& \implies & f(3) = 26 \\[6pt]
\end{array} \nonumber \] - \[ \begin{array}{rcl}
f(x) = 2x^2 + 3x - 1 & \implies & f(-2) = 2(-2)^2 + 3(-2) - 1 \\[6pt]
& \implies & f(-2) = 2(4) + 3(-2) - 1 \\[6pt]
& \implies & f(-2) = 8 - 6 - 1 \\[6pt]
& \implies & f(-2) = 1 \\[6pt]
\end{array} \nonumber \] - \[ \begin{array}{rcl}
f(x) = 2x^2 + 3x - 1 & \implies & f(a) = 2(a)^2 + 3(a) - 1 \\[6pt]
& \implies & f(a) = 2(a^2) + 3(a) - 1 \\[6pt]
& \implies & f(a) = 2a^2 + 3a - 1 \\[6pt]
\end{array} \nonumber \]
- \[ \begin{array}{rcl}
For the function \(f(x)=3x^2−2x+1\), evaluate the function.
- \(f(3)\)
- \(f(−1)\)
- \(f(t)\)
- Answers
-
- \(f(3)=22\)
- \(f(−1)=6\)
- \(f(t)=3t^2−2t−1\)
For the function \(g(x)=3x−5\), evaluate the function.
- \(g(h^2)\)
- \(g(x+2)\)
- \(g(x)+g(2)\)
- Solutions
-
- \[ \begin{array}{rcl}
g(x) = 3x - 5 & \implies & g(h^2) = 3(h^2) - 5 \\[6pt]
& \implies & g(h^2) = 3h^2 - 5 \\[6pt]
\end{array} \nonumber \] - \[ \begin{array}{rcl}
g(x) = 3x - 5 & \implies & g(x + 2) = 3(x + 2) - 5 \\[6pt]
& \implies & g(x + 2) = 3(x) + 3(2) - 5 \\[6pt]
& \implies & g(x + 2) = 3x + 6 - 5 \\[6pt]
& \implies & g(x + 2) = 3x + 1 \\[6pt]
\end{array} \nonumber \] - \[ \begin{array}{rcl}
g(x) + g(2) & = & 3(x) - 5 + \left( 3(2) - 5 \right) \\[6pt]
& = & 3x - 5 + \left( 6 - 5 \right) \\[6pt]
& = & 3x - 5 + \left( 1 \right) \\[6pt]
& = & 3x - 5 + 1 \\[6pt]
& = & 3x - 4 \\[6pt]
\end{array} \nonumber \
- \[ \begin{array}{rcl}
For the function \(g(x)=4x−7\), evaluate the function.
- \(g(m^2)\)
- \(g(x−3)\)
- \(g(x)−g(3)\)
- Answers
-
- \(4m^2−7\)
- \(4x−19\)
- \(x−12\)
The number of unread emails in Sylvia’s account is 75. This number grows by 10 unread emails a day. The function \(N(t)=75+10t\) represents the relation between the number of emails, \(N\), and the time, \(t\), measured in days.
- Determine the independent and dependent variable.
- Find \(N(5)\). Explain what this result means.
- Solutions
-
- The number of unread emails is a function of the number of days. The number of unread emails, \(N\), depends on the number of days, \(t\). Therefore, the variable \(N\) is the dependent variable and the variable \(t\) is the independent variable.
- \[ \begin{array}{rcl}
N(t) = 75 + 10t & \implies & N(5) = 75 + 10(5) \\[6pt]
& \implies & N(5) = 75 + 50 \\[6pt]
& \implies & N(5) = 125 \\[6pt]
\end{array} \nonumber \]Since 5 is the number of days, \(N(5)\), is the number of unread emails after 5 days. After 5 days, there are 125 unread emails in the account.
The number of unread emails in Bryan’s account is 100. This number grows by 15 unread emails a day. The function \(N(t)=100+15t\) represents the relation between the number of emails, \(N\), and the time, \(t\), measured in days.
- Determine the independent and dependent variable.
- Find \(N(7)\). Explain what this result means.
- Answers
-
- \(t\) is the independent variable and \(N\) is the dependent variable.
- 205; the number of unread emails in Bryan’s account on the seventh day.
Key Concepts
- Function Notation: For the function \(y=f(x)\)
- \(f\) is the name of the function
- \(x\) is the domain value
- \(f(x)\) is the range value \(y\) corresponding to the value \(x\).
- We read \(f(x)\) as \(f\) of \(x\) or the value of \(f\) at \(x\).
- Independent and Dependent Variables: For the function \(y=f(x)\),
- \(x\) is the independent variable as it can be any value in the domain
- \(y\) is the dependent variable as its value depends on \(x\)