Functions Question 713
Question: Let g:R→R be defined by g(x) = $\frac{(x^2 + 1)}{(x^2 + 4)}$. Which of the following statements is true about the function g?
Options:
A) one-to-one and onto
B) one-to-one but not onto
C) onto but not one-to-one
D) neither one-to-one nor onto
Show Answer
Answer:
Correct Answer: D
Solution:
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First, let’s look at the behavior of the function: g(x) = $\frac{(x^2 + 1)}{ (x^2 + 4)}$
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Is it one-to-one (injective)? To be one-to-one, each element in the codomain should be paired with at most one element in the domain.
Notice that g(-x) = g(x) for all x: g(-x) = $\frac{((-x)^2 + 1)}{((-x)^2 + 4)} = \frac{(x^2 + 1)}{(x^2 + 4)}$ = g(x)
This means that for every y in the range, there are two x values that map to it (except when x = 0). Therefore, g is not one-to-one.
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Is it onto (surjective)? To determine this, let’s find the range of g.
As x approaches ±∞, g(x) approaches 1: lim(x→±∞) $\frac{(x^2 + 1)}{ (x^2 + 4)}$ = 1
When x = 0, g(0) = 1/4
For all x, $x^2 + 1 < x^2 + 4,$ so g(x) < 1 for all x.
Therefore, the range of g is [1/4, 1).
Since g doesn’t take on all values in R (its codomain), it is not onto.
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Combining our findings: g is neither one-to-one nor onto.
Therefore, the correct answer is D) neither one-to-one nor onto.
BETA
