Onto function

Step 1: (\tan x) is defined on this open interval. Step 2: As (x) approaches the ends, the values cover all real numbers from \(-\infty\) to \(\infty\). Step 3: The chosen interval is very important for trigonometric functions.

Step 1: (\frac{x}{1+|x|}) never reaches (1) or (-1). Step 2: Its range is ((-1,1)), while the codomain is (\mathbb{R}). Step 3: For rational-type expressions, estimate bounds and possible values.

Step 1: The actual range of this function is ((-1,1)). Step 2: The codomain is also ((-1,1)), so no target value is left out. Step 3: A bounded-range function can be onto if the codomain is chosen correctly.

Step 1: (\frac{x^2}{1+x^2}\ge 0) and it remains less than (1). Step 2: At (x=0), (0) is obtained, but (1) is never obtained for any real (x). Step 3: In closed codomains, always check whether endpoints are actually attained.

Step 1: The function starts at (0) and approaches (1), but never becomes (1). Step 2: Hence the actual range is ([0,1)). Step 3: Open or closed endpoints in the codomain can change the answer.

Step 1: \(x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\). Step 2: The minimum value is \(-\frac{1}{4}\), so smaller values are not obtained. Step 3: For quadratics, use the vertex form to find the range.

Step 1: \(f(x)=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\). Step 2: Its range is \(\left[-\frac{1}{4},\infty\right)\), exactly the given codomain. Step 3: Once range and codomain match, onto is proved.

Step 1: (2x) is always an even integer. Step 2: The codomain (\mathbb{Z}) also contains odd integers such as (1) and (3). Step 3: For integer functions, pay attention to parity.

Step 1: (2\mathbb{Z}) is the set of all even integers. Step 2: Any even integer (y) can be written as (y=2k), and taking (x=k) gives (f(x)=y). Step 3: With codomain restricted to even integers, this function is onto.

Step 1: (2n-1) always gives an odd natural number. Step 2: The codomain (\mathbb{N}) also contains even values (2,4,6,\ldots), which are not obtained. Step 3: When a formula gives only odd or even values, check onto carefully.

Step 1: (2n-1) gives all odd natural numbers. Step 2: The codomain is also the set of odd natural numbers. Step 3: If the target set is exactly what the formula produces, the function is onto.

Step 1: Since (1+x^2\ge 1), (\frac{1}{1+x^2}) is greater than (0) and at most (1). Step 2: Its range is ((0,1]), while the codomain is (\mathbb{R}). Step 3: For rational functions, the minimum denominator helps determine the range.

Step 1: At (x=0), the value (1) is obtained. Step 2: As (x) grows large, the value approaches (0) but never becomes (0), so the range is ((0,1]). Step 3: Whether a boundary value is attained matters in onto questions.

Step 1: (\lfloor x\rfloor) gives the greatest integer not greater than (x). Step 2: Hence its range is (\mathbb{Z}), not (\mathbb{R}). Step 3: For the floor function, decimal target values are never obtained.

Step 1: For any integer (k), taking (x=k) gives (\lfloor x\rfloor=k). Step 2: Thus every element of (\mathbb{Z}) is obtained. Step 3: The floor function is onto when the codomain is (\mathbb{Z}).

Step 1: In an onto function, every element of (B) needs at least one preimage. Step 2: (A) has (3) elements while (B) has (4), so all four elements cannot be covered. Step 3: For finite sets, if (|A|<|B|), an onto function cannot exist.

Step 1: For onto, the domain must have at least as many elements as the codomain. Step 2: Here (5\ge 3), so such a function can be constructed. Step 3: Multiple domain elements may map to the same codomain element.

Step 1: Total functions are (2^3=8). Step 2: Non-onto functions use only one codomain element, so there are (2) such functions. Step 3: Therefore onto functions are (8-2=6); complement counting is quick for small sets.

Step 1: Both sets have (2) elements. Step 2: To be onto, both target elements must be hit, so the function is a one-to-one pairing. Step 3: The number of such pairings is (2!=2).

Step 1: In an onto function, range equals codomain. Step 2: Since (B) has (4) elements, the range must also have (4) elements. Step 3: For an onto function, the range size cannot be less than the codomain size.

Step 1: Since (1+x^2\ge 1), (\ln(1+x^2)\ge 0). Step 2: The codomain (\mathbb{R}) contains negative values that are not obtained. Step 3: For logarithmic functions, check both the input expression and the minimum value.

Step 1: (f(x)\ge 0), and (0) is obtained at (x=0). Step 2: For any (y\ge 0), from (1+x^2=e^y), we can take (x=\sqrt{e^y-1}). Step 3: Solving for (x) from the target value is a strong method to prove onto.

Step 1: (x^3-3x^2+3x=(x-1)^3+1). Step 2: The cubic form ((x-1)^3+1) takes all real values. Step 3: Rewriting into cubic form makes the range clear.

Step 1: Since (g) is onto, for every target (y), there is some (u) with (g(u)=y). Step 2: Since (f) is onto, for that (u), there is some (x) with (f(x)=u). Step 3: Therefore (g(f(x))=y), so the composition is also onto.

Step 1: When (x\in[-2,2]), (x^2) takes values from (0) to (4). Step 2: For any (y\in[0,4]), taking (x=\sqrt{y}) gives (x\in[-2,2]) and (f(x)=y). Step 3: In exams, decide onto only after checking both the domain and the codomain.