Question
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  1. The function and are defined as

    a. State the range of
    b. Find the domain of
    c. Sketch the graph of indicating the intercepts with the -axis
    d. State with a reason, whether is injective or not

Ask by Warren Schofield. in Zimbabwe
Jan 16,2025

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Tutor-Verified Answer

Answer

The function is not injective because different inputs can produce the same output.

Solution

d. Determining Whether is Injective
A function is injective (or one-to-one) if every element of the function’s range is mapped by exactly one element of its domain. In other words, for to be injective, whenever , it must follow that .
Let’s analyze the function :
  1. Symmetry Analysis:
    • The function is a symmetric function about the y-axis because for all .
  2. Testing Injectivity:
    • Consider two different inputs: and where .
    • Compute and .
    • Here, even though .
  3. Conclusion:
    • Since there exist distinct values and such that , the function is not injective.
Reason: The function is symmetric about the y-axis, meaning that for any . Therefore, distinct inputs and produce the same output, violating the condition for injectivity.
Answer Summary:
Problem 4.d Answer:
The function is not injective because for any real number , meaning distinct inputs can produce the same output.

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The function is a downward-opening parabola with its vertex at . The range of is since as moves away from 0, decreases without bound.
For the function to be defined, must be non-negative. Thus, we need to find when , which simplifies to , or . Therefore, the domain of is the interval .
To sketch , we first identify that touches the x-axis when , which occurs at . Thus, the intercepts are at points and . The function will reflect the negative part of across the x-axis, so between the intercepts, the graph will rise to a maximum of 4 at and then drop back to 0 on either side.
The function is not injective (or one-to-one) because it fails the horizontal line test. For example, both and yield a value of 3 (i.e., ). This means there are multiple values that produce the same output, confirming that is not injective.

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