The function f:(−∞,∞)→(−∞,1), defined by f(x)=2x+2−x2x−2−x, is:
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Step-by-Step Solution
Step 1: Simplify the function
First, we rewrite 2−x as 2x1 to simplify the expression. This makes it easier to combine the terms in the numerator and the denominator.
Step 2: Combine terms
Next, we find a common denominator for the terms in the numerator and the denominator. After combining them, we can cancel out the common factor of 2x from both the numerator and the denominator, resulting in a simpler form of the function.
Step 3: Analyze the range of the function
To understand the range of the function, we can rewrite it by adding and subtracting 1 in the numerator. This transformation allows us to express the function as 1−22x+12, which makes it easier to analyze its behavior.
Step 4: Determine the behavior of 22x
The term 22x is always positive for any real value of x. As x→∞, 22x→∞. As x→−∞, 22x→0.
Step 5: Determine the range of f(x)
When x approaches infinity, 22x+1 approaches infinity, causing the fraction 22x+12 to approach 0. Thus, f(x) approaches 1−0=1. When x approaches negative infinity, 22x+1 approaches 1, causing the fraction 22x+12 to approach 2. Thus, f(x) approaches 1−2=−1. Since 22x+1 is always greater than 1, the fraction 22x+12 is always positive and less than 2. Therefore, f(x) is always between -1 and 1.
Step 6: Check for injectivity (one-to-one)
To check if the function is injective (one-to-one), we assume f(x1)=f(x2) and show that this implies x1=x2. After cross-multiplication and simplification, we find that 22x1=22x2, which means 2x1=2x2, and therefore x1=x2. This confirms that the function is injective.
Step 7: Check for surjectivity (onto)
To check for surjectivity (onto), we need to see if for every y in the codomain (−∞,1), there exists an x in the domain (−∞,∞) such that f(x)=y. We solve for x in terms of y. The expression for x is defined only when 1−yy+1>0. This inequality holds when −1<y<1. Since the given codomain is (−∞,1), and our range is (−1,1), the function is not surjective onto (−∞,1). However, if the codomain was (−1,1), it would be surjective.