Onto function

Step 1: For any (y\ne 0), from (y=\frac{1}{x}), we get (x=\frac{1}{y}). Step 2: This (x) is also non-zero, so it lies in the domain. Hence the function is onto. Different (x) values give different reciprocals, so it is one-one too. Step 3: For reciprocal functions, removing zero from domain and codomain is essential.

Step 1: Since (1+x^2\ge 1), (\frac{1}{1+x^2}\le 1) and is always positive. Step 2: Its range is ((0,1]), while the codomain is (\mathbb{R}). So many real numbers such as (2) and (-1) are not obtained. Step 3: Always identify the exact range.

Step 1: The range of this function is ((0,1]). Step 2: The codomain is also ((0,1]), so the function is onto. But (f(1)=f(-1)=\frac{1}{2}), so it is not one-one. Step 3: For even functions, compare values at (x) and (-x).

Step 1: From (y=3x^3-2), we get (x=\sqrt[3]{\frac{y+2}{3}}). Step 2: This is real for every real (y), so the function is onto. The cubic function is increasing, so it is one-one too. Step 3: Cube roots are defined for every real number.

Step 1: Put (y=x-1), so (x=y+1). Step 2: If (y\ge 0), then (x\ge 1), so (x) lies in the domain. Hence every codomain element is an image. Step 3: For interval-based functions, always check whether the obtained (x) lies in the domain.

Step 1: On ([0,2]), the range of (x^2) is ([0,4]). Step 2: This equals the codomain, so the function is onto. On ([0,2]), (x^2) is increasing, so different (x) values give different images. Step 3: On a closed interval, endpoints often help find the range quickly.

Step 1: On ([0,2]), the maximum value of (x^2) is (4). Step 2: The codomain is ([0,5]), but elements like (5) are not obtained. Thus the range ([0,4]) is smaller than the codomain. Step 3: Identify the extra part of the codomain in onto questions.

Step 1: ((x-2)^2\ge 0), so the minimum value is (3). Step 2: The range is ([3,\infty)), equal to the codomain. Hence the function is onto. But (x=1) and (x=3) give the same value, so it is not one-one. Step 3: The vertex of a quadratic tells the range.

Step 1: Since (x^2\ge 0), (4-x^2\le 4). Step 2: For large (x), values become very negative, so the range is ((-\infty,4]). This equals the codomain. Step 3: For a downward-opening quadratic, check the maximum value.

Step 1: (|x-4|\ge 0), so (f(x)\ge 1). Step 2: For every (y\ge 1), the equation (|x-4|=y-1) has a solution, so the function is onto. But (x=3) and (x=5) give the same value. Step 3: A modulus function can be onto but often not one-one.

Step 1: In the given domain, (x\ge 4), so (|x-4|=x-4). Step 2: The function becomes (f(x)=x-3), whose range is ([1,\infty)). This equals the codomain and is increasing. Step 3: Before removing modulus, check the domain condition.

Step 1: (x^2-4x+5=(x-2)^2+1). Step 2: So the minimum value is (1) and the range is ([1,\infty)). The codomain (\mathbb{R}) contains (0) and negative values, which are not obtained. Step 3: Completing the square is very useful for finding range.

Step 1: (x^3+3x) is increasing everywhere, because as (x) increases, the value increases. Step 2: As (x\to\infty), the value goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Hence every real (y) is obtained and the function is onto. Step 3: An increasing cubic is often both onto and one-one.

Step 1: The codomain contains only even integers. Step 2: Any even integer (y) can be written as (y=2k), where (k\in\mathbb{Z}). Then (f(k)=y), so the function is onto. Different (n) values give different (2n). Step 3: Changing the codomain can make the same rule onto.

Step 1: (2n-1) always gives an odd natural number. Step 2: The codomain (\mathbb{N}) contains even natural numbers such as (2,4,6), which are not images. Therefore the function is not onto. Step 3: Identify the type of values produced by the formula.

Step 1: The codomain is the set of all odd natural numbers. Step 2: Every odd number can be written as (y=2k-1), so (k) is its preimage. Different (n) values give different odd numbers. Step 3: When the codomain exactly equals the range, the function is onto.

Step 1: Since (e^x>0), (e^x+1>1). Step 2: For any (y>1), (x=\ln(y-1)), which is real. Hence the function is onto. It is increasing, so it is also one-one. Step 3: Vertical shifts change the range of exponential functions.

Step 1: (\frac{x^2}{1+x^2}) is always at least (0) and less than (1). Step 2: At (x=0), we get (0), but (1) is not obtained for any real (x). So the range is ([0,1)), not the codomain ([0,1]). Step 3: Distinguish between a limiting value and an attained value.

Step 1: The range of this function is ([0,1)). Step 2: The codomain is also ([0,1)), so the function is onto. But (f(2)=f(-2)), so it is not one-one. Step 3: With the correct codomain, the same function can become onto.

Step 1: This is a continuous polynomial with an odd leading degree. Step 2: As (x\to\infty), the value goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Hence it takes every real value in between and is onto. Step 3: Use the end behavior of continuous cubic polynomials to judge onto nature.

Step 1: On ([-1,1]), the range of (x^2) is ([0,1]). Step 2: This equals the given codomain, so the function is onto. But (f(-1)=f(1)=1), so it is not one-one. Step 3: For even functions, compare values at (x) and (-x).

Step 1: On ([-1,1]), the greatest value of (x^2) is (1). Step 2: So the range is ([0,1]), while the codomain is ([0,2]). The codomain element (2) is not obtained from any (x). Step 3: While checking onto, observe maximum and minimum values carefully.

Step 1: The value of this function always lies between (-1) and (1). Step 2: For any (y\in(-1,1)), (x=\frac{y}{\sqrt{1-y^2}}) is real, so every codomain element has a preimage. The function is also increasing, so it is one-one. Step 3: For fractional square-root functions, first identify the limiting range.

Step 1: The value of (\frac{x}{\sqrt{1+x^2}}) remains greater than (-1) and less than (1). Step 2: For very large (x), the value approaches (1), but never becomes (1). Similarly, (-1) is not obtained. Step 3: Approaching a limiting value and attaining it are different ideas.

Step 1: When an integer is divided by (3), the remainder is one of (0,1,2). Step 2: All three remainders are obtained, for example (f(3)=0), (f(4)=1), and (f(5)=2). Hence the function is onto. But many integers give the same remainder, so it is not one-one. Step 3: In remainder-based questions, list all possible remainders.