(a) Given the function [tex]$f(x)=3|x-1|-2$[/tex].
(i) Sketch the graph of the function.
(ii) Find the values of [tex]$x$[/tex] for which [tex]$f(x) \geq 0$[/tex]
(iii) Solve the equation [tex]$f(x)=1$[/tex].
Answer (2)
The graph of f ( x ) = 3∣ x − 1∣ − 2 is a V-shape with a vertex at (1, -2). The function is non-negative for x ≤ 3 1 and x ≥ 3 5 , with solutions to f ( x ) = 1 being x = 0 and x = 2 .
;
Analyze the absolute value function by considering two cases: x ≥ 1 and x < 1 , leading to f ( x ) = 3 x − 5 and f ( x ) = 1 − 3 x , respectively.
Solve the inequality f ( x ) ≥ 0 , which is equivalent to ∣ x − 1∣ ≥ 3 2 , resulting in x ≤ 3 1 or x ≥ 3 5 .
Solve the equation f ( x ) = 1 , which is equivalent to ∣ x − 1∣ = 1 , giving the solutions x = 0 and x = 2 .
The values of x for which f ( x ) = 1 are 0 , 2 .
Explanation
Understanding the Problem We are given the function f ( x ) = 3∣ x − 1∣ − 2 . Our goal is to sketch its graph, find the values of x for which =" 0, and solve the equation"> f ( x ) " >= "0 , an d so l v e t h ee q u a t i o n f(x) = 1$.
Analyzing the Absolute Value To sketch the graph of f ( x ) , we need to consider two cases based on the absolute value: when x ≥ 1 and when x < 1 .
Case 1: x ≥ 1 Case 1: x ≥ 1 . In this case, ∣ x − 1∣ = x − 1 , so f ( x ) = 3 ( x − 1 ) − 2 = 3 x − 3 − 2 = 3 x − 5 . This is a linear function with a slope of 3 and a y-intercept of -5. However, we only consider the part of the line where x ≥ 1 . When x = 1 , f ( 1 ) = 3 ( 1 ) − 5 = − 2 .
Case 2: x < 1 Case 2: x < 1 . In this case, ∣ x − 1∣ = − ( x − 1 ) = 1 − x , so f ( x ) = 3 ( 1 − x ) − 2 = 3 − 3 x − 2 = 1 − 3 x . This is a linear function with a slope of -3 and a y-intercept of 1. However, we only consider the part of the line where x < 1 . When x = 1 , f ( 1 ) = 1 − 3 ( 1 ) = − 2 .
Finding when f(x) ≥ 0 Now, let's find the values of x for which f ( x ) ≥ 0 . This means 3∣ x − 1∣ − 2 ≥ 0 , which simplifies to ∣ x − 1∣ ≥ 3 2 . This inequality holds if x − 1 ≥ 3 2 or x − 1 ≤ − 3 2 .
Solving the Inequality Solving x − 1 ≥ 3 2 , we get x ≥ 1 + 3 2 = 3 5 . Solving x − 1 ≤ − 3 2 , we get x ≤ 1 − 3 2 = 3 1 . Therefore, f ( x ) ≥ 0 when x ≤ 3 1 or x ≥ 3 5 .
Solving f(x) = 1 Finally, let's solve the equation f ( x ) = 1 . This means 3∣ x − 1∣ − 2 = 1 , which simplifies to 3∣ x − 1∣ = 3 , and thus ∣ x − 1∣ = 1 . This equation holds if x − 1 = 1 or x − 1 = − 1 .
Finding the Solutions Solving x − 1 = 1 , we get x = 2 . Solving x − 1 = − 1 , we get x = 0 . Therefore, the solutions to f ( x ) = 1 are x = 0 and x = 2 .
Examples
Absolute value functions are useful in many real-world scenarios. For example, consider a thermostat that is set to 70 degrees Fahrenheit. The function f ( x ) = ∣ x − 70∣ represents the difference between the actual temperature x and the set temperature. If you want to ensure the temperature stays within 5 degrees of the set point, you would solve the inequality ∣ x − 70∣ ≤ 5 . This type of problem arises in various engineering and scientific applications where maintaining a certain tolerance is crucial.
