u use scratch pap -2x - 3, x < -2 0 < x< 3 0, x > 3 the function f (x) = }-x² + 4, %3D 1. Evaluate f(5): S 2. Evaluate f(1): 3. Evaluate f(-7): 4. Evaluate f (0): 5. Find the domain of f (x) a. (-00, 00) 6. (-0,-2)U(0 c. (-2,3) d. (-0,-2) U (0, co) 0, 00
u use scratch pap -2x - 3, x < -2 0 < x< 3 0, x > 3 the function f (x) = }-x² + 4, %3D 1. Evaluate f(5): S 2. Evaluate f(1): 3. Evaluate f(-7): 4. Evaluate f (0): 5. Find the domain of f (x) a. (-00, 00) 6. (-0,-2)U(0 c. (-2,3) d. (-0,-2) U (0, co) 0, 00
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Related questions
Question
Use the function f(x) and evil are.
![### Section 3 Test
For the function \( f(x) \):
\[
f(x) =
\begin{cases}
-2x - 3, & x < -2 \\
-x^2 + 4, & 0 \leq x \leq 3 \\
0, & x > 3
\end{cases}
\]
1. Evaluate \( f(5) \):
2. Evaluate \( f(1) \):
3. Evaluate \( f(-7) \):
4. Evaluate \( f(0) \):
5. Find the domain of \( f(x) \)
a. \((-\infty, \infty)\)
b. \((-\infty, -2) \cup [0, \infty)\)
c. \((-2, 3)\)
d. \((-\infty, -2) \cup (0, \infty)\)
(Note: Option "b" is circled, indicating it might be the chosen answer.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9984d7ae-1a86-419a-b4b6-309f0e42346f%2F22438aa9-8eac-49d3-b393-ac1d1f7f27ec%2Filk2d6c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Section 3 Test
For the function \( f(x) \):
\[
f(x) =
\begin{cases}
-2x - 3, & x < -2 \\
-x^2 + 4, & 0 \leq x \leq 3 \\
0, & x > 3
\end{cases}
\]
1. Evaluate \( f(5) \):
2. Evaluate \( f(1) \):
3. Evaluate \( f(-7) \):
4. Evaluate \( f(0) \):
5. Find the domain of \( f(x) \)
a. \((-\infty, \infty)\)
b. \((-\infty, -2) \cup [0, \infty)\)
c. \((-2, 3)\)
d. \((-\infty, -2) \cup (0, \infty)\)
(Note: Option "b" is circled, indicating it might be the chosen answer.)
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