Chapter 2: Problem 50
For each piecewise function, find the specified function values. $$\begin{aligned} &f(x)=\left\\{\begin{array}{ll} -5 x-8, & \text { for } x<-2 \\ \frac{1}{2} x+5, & \text { for }-2 \leq x \leq 4 \\ 10-2 x, & \text { for } x>4 \end{array}\right.\\\ &f(-4), f(-2), f(4), \text { and } f(6) \end{aligned}$$
Short Answer
Step by step solution
Identify the appropriate function for f(-4)
Calculate f(-4)
Identify the appropriate function for f(-2)
Calculate f(-2)
Identify the appropriate function for f(4)
Calculate f(4)
Identify the appropriate function for f(6)
Calculate f(6)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function Evaluation
We can illustrate this with an exercise: If we have a piecewise function like this:
\[f(x)=\left\{\begin{array}{ll}-5 x-8, & \text { for } x<-2 \frac{1}{2} x+5, & \text { for }-2 \leq x \leq 4 \10-2 x, & \text { for } x>4\end{array}\right.\]Let's compute \( f(-4), f(-2), f(4), \text{ and } f(6) \).For \( f(-4) \): Since \( -4 < -2 \), we use \( -5x-8 \). Substituting \( -4 \) into the function: \( f(-4) = -5(-4) - 8 = 20 - 8 = 12 \).
For \( f(-2) \): Since \( -2 \leq x \leq 4 \), we use \( \frac{1}{2}x + 5 \). Substituting \( -2 \) into the function: \( f(-2) = \frac{1}{2}(-2) + 5 = -1 + 5 = 4 \).
For \( f(4) \): Since \( -2 \leq x \leq 4 \), we use \( \frac{1}{2}x + 5 \). Substituting \( 4 \) into the function: \( f(4) = \frac{1}{2}(4) + 5 = 2 + 5 = 7 \).
For \( f(6) \): Since \( x > 4 \), we use \( 10-2x \). Substituting \( 6 \) into the function: \( f(6) = 10 - 2(6) = 10 - 12 = -2 \).
Domain and Range
The range of the function is the set of possible output values (y-values) that the function can produce.
Let's apply this to our piecewise function example.
\[f(x)=\left\{\begin{array}{ll}-5 x-8, & \text { for } x<-2 \frac{1}{2} x+5, & \text { for }-2 \leq x \leq 4 \10-2 x, & \text { for } x>4\end{array}\right.\]
The domain here covers all real numbers because there are no restrictions on \( x \). The specific segments are:
- \( x < -2 \)
- \( -2 \leq x \leq 4 \)
- \( x > 4 \)
The ranges depend on each segment.
- For \( x < -2 \), solving \( -5x-8 \) gives us values from \( -fty \text{ to } fty \).
- For \( -2 \leq x \leq 4 \), solving \( \frac{1}{2}x + 5 \) gives values from \( -1 \text{ to } 7 \).
- For \( x > 4 \), values of \( 10 - 2x \) range from values greater \( -fty \text{ to } ewline \) as \( x \rightarrow fty \) .
Substitution Method
For our example:
\[f(x)=\left\{\begin{array}{ll}-5 x-8, & \text { for } x<-2 \frac{1}{2} x+5, & \text { for }-2 \leq x \leq 4 \10-2 x, & \text { for } x>4\end{array}\right.\]
This means:
- If \( x = -4 \), substitute -4 into \( -5x-8 \) to get 12.
- If \( x = -2 \), substitute -2 into \( \frac{1}{2}x + 5 \) to get 4.
- If \( x = 4 \), substitute 4 into \( \frac{1}{2}x + 5 \) to get 7.
- If \( x = 6 \), substitute 6 into \( 10 - 2x \) to get -2.