a. Since 0 is less than 2, to evaluate f(0) we substitute x = 0 using the first equation, Hence, f(0) = (0)² – (0) + 4 : f(0) = 4 %3D %3D b. To evaluate f(2), substitute x = 2 using the second equation, f(x) = x – 2. Hence, f(2) = (2) – 2 * f(2) = 0 %3D c. To evaluate f(5), substitute x = 5 using the second equation. Hence,

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,...
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a. Since 0 is less than 2, to evaluate f(0) we substitute x = 0 using the first equation, f(x) = x² – x +
Hence,
f(0) = (0)² – (0) + 4
i f(0) = 4
%3D
b. To evaluate f(2), substitute x = 2 using the second equation, f(x) = x – 2. Hence,
f(2) = (2) – 2
f(2) = 0
%3D
%3D
c. To evaluate f(5), substitute x = 5 using the second equation. Hence,
f(5) = (5) – 2
* f(5) = 3
%3D
Transcribed Image Text:a. Since 0 is less than 2, to evaluate f(0) we substitute x = 0 using the first equation, f(x) = x² – x + Hence, f(0) = (0)² – (0) + 4 i f(0) = 4 %3D b. To evaluate f(2), substitute x = 2 using the second equation, f(x) = x – 2. Hence, f(2) = (2) – 2 f(2) = 0 %3D %3D c. To evaluate f(5), substitute x = 5 using the second equation. Hence, f(5) = (5) – 2 * f(5) = 3 %3D
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