Onto function

Step 1: For onto, every element of the codomain must have a preimage. Step 2: From (y=x^3+1), (x=\sqrt[3]{y-1}), which is real for every real (y). The function is also increasing, so it is one-one. Step 3: In exams, solve (y=f(x)) for (x).

Step 1: Both elements (a) and (b) of the codomain appear as images. Step 2: Hence the function is onto. But (2) and (3) have the same image (b), so it is not one-one. Step 3: For finite sets, check whether every codomain element appears in the image list.

Step 1: For any integer (y), put (y=n+5). Step 2: Then (n=y-5), which is also an integer. So every (y) has a preimage and the function is onto. Different (n) values also give different images. Step 3: Addition or subtraction functions on integers are often onto.

Step 1: If (\mathbb{N}={1,2,3,\ldots}), the smallest value of (n+1) is (2). Step 2: The codomain element (1) is not obtained from any (n). Hence the range is not equal to the codomain. Step 3: For natural numbers, check the smallest possible value carefully.

Step 1: To check onto, write (y=2x-7). Step 2: We get (x=\frac{y+7}{2}), which is real for every real (y). Thus every codomain element is obtained. Step 3: A linear function with non-zero slope is onto from (\mathbb{R}) to (\mathbb{R}).

Step 1: In a constant function, every (x) has the image (5). Step 2: The range is ({5}), while the codomain is (\mathbb{R}). Other codomain elements such as (0) and (2) are not obtained. Step 3: A constant function is usually not onto over a large codomain.

Step 1: The range of (|x|) is ([0,\infty)). Step 2: This equals the given codomain, so the function is onto. But (|2|=|-2|), so it is not one-one. Step 3: For modulus functions, compare positive and negative input pairs.

Step 1: (|x|) can never be negative. Step 2: The codomain (\mathbb{R}) includes negative numbers such as (-3). The equation (|x|=-3) has no real solution. Step 3: Remember: if the range is smaller than the codomain, the function is not onto.

Step 1: On ([0,\infty)), (x^2) is increasing. Step 2: For any (y\in[0,\infty)), (x=\sqrt{y}) is in the domain, so the function is onto. Different (x) values give different squares, so it is one-one too. Step 3: Changing the domain can change the nature of a function.

Step 1: Since (x^2\ge 0), (x^2-1\ge -1). Step 2: For any (y\ge -1), we can take (x=\sqrt{y+1}). Hence every codomain element has a preimage. Step 3: For quadratics, find the minimum value to get the range quickly.

Step 1: (e^x) is always positive and its range is ((0,\infty)). Step 2: For every (y>0) in the codomain, (x=\ln y) is real. So the function is onto and also one-one because it is increasing. Step 3: Remember the range of exponential functions.

Step 1: (e^x>0) for every real (x). Step 2: The codomain (\mathbb{R}) contains (0) and negative numbers, which are not obtained from (e^x). Hence the range is smaller than the codomain. Step 3: Reading the codomain carefully is most important in onto questions.

Step 1: The range of (\ln x) is (\mathbb{R}). Step 2: For any real (y), (x=e^y>0), which lies in the domain. So the function is onto. Since (\ln x) is increasing, it is also one-one. Step 3: Remember the domain-range link between (\ln x) and (e^x).

Step 1: The range of (\sin x) is ([-1,1]). Step 2: The codomain is also ([-1,1]), so the function is onto. But (\sin 0=\sin \pi=0), so it is not one-one. Step 3: Periodicity often breaks one-one nature in trigonometric functions.

Step 1: The range of (\cos x) is ([-1,1]). Step 2: The codomain is the same, so every codomain element is the image of some (x). Hence the function is onto. Step 3: For onto, the range must equal the codomain.

Step 1: On this interval, (\tan x) is strictly increasing. Step 2: As (x) approaches \(-\frac{\pi}{2}\), the value becomes very small, and near \(\frac{\pi}{2}\), it becomes very large. Hence the range is \(\mathbb{R}\). Step 3: Choosing a suitable interval can make a trigonometric function onto and one-one.

Step 1: (x^3-x) is a continuous cubic polynomial. Step 2: As (x\to\infty), its value goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Therefore every real (y) is obtained. Step 3: Polynomials with odd leading degree are often onto from (\mathbb{R}) to (\mathbb{R}).

Step 1: (x^4\ge 0), so (x^4+1\ge 1). Step 2: The codomain is (\mathbb{R}), but (0) and negative numbers are not images. Therefore it is not onto. Step 3: For even-degree polynomials, check the minimum value.

Step 1: All three elements (p,q,r) of (B) appear at least once as images. Step 2: Hence the function is onto. But (2) and (3) both map to (q), so it is not one-one. Step 3: For finite functions, mark all elements of the codomain while checking onto.

Step 1: In an onto function, every codomain element must be covered by some domain element. Step 2: Here the domain has (3) elements and the codomain has (4) elements. Three elements cannot cover four distinct codomain elements. Step 3: For finite sets, onto requires (|A|\ge |B|).

Step 1: For an onto function (A\to B) between finite sets, (A) must have at least as many elements as (B). Step 2: Here (5\ge 3), so an onto function can be made. But not every function must be onto. Step 3: Distinguish between existence and all-function property.

Step 1: The total number of functions is (2^4=16). Step 2: A function is not onto if all elements map only to (a) or only to (b), giving (2) such functions. Hence onto functions are (16-2=14). Step 3: For a two-element codomain, subtract the two constant functions from the total.

Step 1: Total functions are (2^3=8). Step 2: Only two functions are not onto: all elements map to (a), or all map to (b). Hence onto functions are (8-2=6). Step 3: In small finite sets, first count total functions.

Step 1: However large (x) is, (\frac{x}{1+|x|}) remains less than (1). In the negative direction, it remains greater than (-1). Step 2: So the range is ((-1,1)), while the codomain is (\mathbb{R}). Step 3: In rational-type functions, limiting values often stay outside the range.

Step 1: Put (y=\frac{x+1}{x-1}). Step 2: Solving gives (x=\frac{y+1}{y-1}). This is defined when (y\ne 1), which is exactly the codomain condition. So every (y\in\mathbb{R}\setminus{1}) has a preimage. Step 3: For rational functions, solve for (x) in terms of (y).