Onto function

Step 1: (f(x)=x^2+2x+1=(x+1)^2). Step 2: The range is ([0,\infty)), exactly the given codomain. Step 3: In exams, compare the range with the codomain to decide onto.

Step 1: (f(x)=(x+1)^2) and the domain has (x\ge -1). Step 2: Since (x+1\ge0), every (y\in[0,\infty)) is obtained. Step 3: A restricted domain can still make a function onto if it covers the whole codomain.

Step 1: (e^x) is always positive. Step 2: For every (y>0), take (x=\ln y), then (e^x=y). Step 3: For exponential functions, the correct codomain is the key to onto checking.

Step 1: (e^x>0) for all real (x). Step 2: (0) is in ([0,\infty)), but there is no (x) such that (e^x=0). Step 3: A closed endpoint in the codomain can change onto status.

Step 1: The value of the function always lies between (-1) and (1). Step 2: For (y\in(-1,1)), taking (x=\frac{y}{1-|y|}) gives (f(x)=y). Step 3: Missing endpoints does not matter when the codomain is open.

Step 1: (\frac{x}{1+|x|}) stays greater than (-1) and less than (1). Step 2: (-1) and (1) are in the codomain but are not obtained for any real (x). Step 3: Always check endpoints when the codomain is closed.

Step 1: (x^3-3x) is a polynomial continuous on all real numbers. Step 2: As (x\to\infty), (f(x)\to\infty), and as (x\to-\infty), (f(x)\to-\infty), so every real value is obtained. Step 3: An odd-degree leading term often helps prove onto from (\mathbb{R}) to (\mathbb{R}).

Step 1: \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\). Step 2: Its minimum value is \(\frac{3}{4}\), so real values below it are missed. Step 3: Completing the square is a useful way to find the range.

Step 1: Completing the square gives \(f(x)=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\). Step 2: The range is \(\left[\frac{3}{4},\infty\right)\), the same as the codomain. Step 3: A function is onto when its range equals its codomain.

Step 1: For any (y\ne1), write (y=\frac{x+1}{x-1}). Step 2: Solving gives (x=\frac{y+1}{y-1}), which is defined for (y\ne1) and also (x\ne1). Step 3: If every target value gives a valid (x), the function is onto.

Step 1: Putting (\frac{x+1}{x-1}=1) gives (x+1=x-1). Step 2: This is impossible, so (1) is never obtained. Step 3: If the codomain is (\mathbb{R}) and one real value is missed, the function is not onto.

Step 1: Since (x^2\ge0), (3-x^2\le3). Step 2: For every (y\le3), take (x=\sqrt{3-y}) or (x=-\sqrt{3-y}). Step 3: When the maximum value is attained, the closed endpoint is included in the range.

Step 1: At (x=0), (f(0)=3). Step 2: But (3) is not in the given codomain ((-\infty,3)), so the function is not well-defined for this codomain. Step 3: Before onto checking, verify that every output lies in the codomain.

Step 1: Let the target value be (m\in\mathbb{Z}). Step 2: (n=m-5) is also an integer and (f(n)=m). Step 3: A linear shift on integers is often onto.

Step 1: If (\mathbb{N}={1,2,3,\ldots}), then (f(n)=n+5\ge6). Step 2: Codomain values such as (1,2,3,4,5) are not obtained. Step 3: For integer and natural number functions, always check the starting values.

Step 1: (2n+1) is always an odd integer. Step 2: The codomain (\mathbb{Z}) also contains even integers like (0) and (2), which are missed. Step 3: For integer functions, parity often gives a quick test.

Step 1: The codomain contains only odd integers. Step 2: For every odd integer (y), (n=\frac{y-1}{2}) is an integer and (f(n)=y). Step 3: Restricting the codomain suitably can make the same function onto.

Step 1: In an onto function, every element of (B) must be an image of some element of (A). Step 2: Here (A) has (3) elements and (B) has (4), so covering all four elements is impossible. Step 3: For finite sets, if (|A|<|B|), an onto function cannot exist.

Step 1: Total functions are (3^5). Step 2: By inclusion-exclusion, onto functions (=3^5-\binom{3}{1}2^5+\binom{3}{2}1^5=243-96+3=150). Step 3: For counting onto functions on finite sets, use inclusion-exclusion.

Step 1: Total functions are (2^4=16). Step 2: Non-onto functions use only one codomain element, so there are (2) such functions. Step 3: Hence onto functions (=16-2=14).

Step 1: For onto, all (p,q,r) must appear among the images. Step 2: In the first option, (p,q,r) all occur. Step 3: In mapping-list questions, form the image set and compare it with the codomain.

Step 1: The range of (\sin x) is ([-1,1]). Step 2: The codomain (\mathbb{R}) contains values such as (2), which are never obtained. Step 3: For trigonometric functions, remembering the range is very important for onto checking.

Step 1: The range of (\sin x) is exactly ([-1,1]). Step 2: The values (-1) and (1) are also obtained at suitable values of (x). Step 3: If the range equals the codomain, the function is onto.

Step 1: On this interval, (\sin x) takes every value from (-1) to (1). Step 2: The codomain is also ([-1,1]), so no value is missed. Step 3: On restricted intervals, continuity and monotonic behavior help identify the range.

Step 1: The open interval does not include (x=\pm\frac{\pi}{2}). Step 2: Hence the range of (\sin x) is ((-1,1)), while the codomain is ([-1,1]). Step 3: The difference between open and closed intervals can decide onto status.