32. Show that f:R→R defined as f(x)=1+x2x is one-one but not onto. (function, domain, codomain, one-to-one, onto)
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Step-by-Step Solution
Step 1: Check for One-one (Injectivity)
To prove that the function f(x) is one-one, we assume that f(x1)=f(x2) for any two elements x1,x2 in the domain R and then show that this implies x1=x2.
Step 2: Set up the equation for one-one
Substitute the function definition into the assumption f(x1)=f(x2). This forms an equation that we will solve for x1 and x2.
Step 3: Square both sides
To eliminate the square roots, we square both sides of the equation. This operation simplifies the expression, removing the square root from the denominator.
Step 4: Rearrange the equation
Cross-multiply the terms to remove the denominators. This step helps in bringing all terms to a single line, making further algebraic manipulation easier.
Step 5: Expand and simplify
Expand the products on both sides and observe that the term x12x22 cancels out from both sides, leading to x12=x22.
Step 6: Analyze x12=x22
From x12=x22, we deduce that x1 must be either equal to x2 or x1 must be equal to −x2. We need to ensure that only x1=x2 is possible for the function to be one-one.
Step 7: Check the sign of f(x)
Notice that f(x) has the same sign as x. This means that f(x1)=f(x2) cannot hold if x1 and x2 have opposite signs (unless both are zero), as one would produce a positive value and the other a negative value.
Step 8: Conclusion for one-one
Since f(x1)=f(x2) implies that f(x1) and f(x2) must have the same sign, x1 and x2 must have the same sign. Therefore, from x1=±x2, the only valid option is x1=x2. This confirms that the function is one-one (injective).
Step 9: Check for Onto (Surjectivity)
To check if the function is onto, we need to determine if for every value y in the codomain R, there exists an x in the domain R such that f(x)=y. We'll try to express x in terms of y to find the range of the function.
Step 10: Square both sides and solve for x2
Square both sides of the equation y=f(x) to eliminate the square root and then algebraically rearrange the equation to isolate x2. This will define the relationship between x and y that must hold.
Step 11: Solve for x
Take the square root of both sides to find x in terms of y. For x to be a real number, the expression under the square root must be non-negative, and the denominator must not be zero.
Step 12: Determine the range restrictions
For x to be defined, the term 1−y2 in the denominator must be strictly positive. This implies that y2 must be less than 1, which means y must be between −1 and 1 (exclusive). If y=±1, the denominator becomes zero, making x undefined. Also, consider the case where y=0, which gives x=0 and is valid.
Step 13: Consider the sign of x and y
From the original function, f(x)=1+x2x, we know that f(x) has the same sign as x. Therefore, y must have the same sign as x. If y>0, we choose the positive square root for x. If y<0, we choose the negative square root for x. This ensures consistency with the original function definition of f(x).
Step 14: Final range determination
Combining the conditions, the range of the function f(x) is the open interval (−1,1). This means that for any y outside this interval (i.e., y≤−1 or y≥1), there is no real x such that f(x)=y. Since the codomain is R and the range is (−1,1), the range is not equal to the codomain.
Step 15: Conclusion for Onto
Since the range of the function, which is the interval (−1,1), is not equal to the codomain, which is R, the function is not onto (surjective). For example, there is no x for which f(x)=2 because 2 is not in the range (−1,1).