Onto function

Step 1: Since (x^2\ge 0), (x^2+2\ge 2). Step 2: For every (y\ge 2), taking (x=\sqrt{y-2}) gives (f(x)=y). Step 3: In exams, first find the range and compare it with the codomain.

Step 1: The value of (f(x)=x^2+2) is always at least (2). Step 2: The codomain (\mathbb{R}) contains (1), but no real (x) satisfies (x^2+2=1). Step 3: For onto checking, one missing codomain element is enough.

Step 1: For (x\ge 0), (f(x)=\frac{x}{1+x}), and for (x<0), (f(x)=\frac{x}{1-x}). Step 2: For any (y\in(-1,1)), choose (x=\frac{y}{1-y}) when (y\ge0), and (x=\frac{y}{1+y}) when (y<0). Step 3: For piecewise-looking expressions, find the preimage according to sign.

Step 1: The range of (e^x) is ((0,\infty)). Step 2: The codomain is also ((0,\infty)), so every positive (y) has preimage (x=\ln y). Step 3: In exponential functions, (0) is not in the range, so read the codomain carefully.

Step 1: (e^x) is always positive. Step 2: The codomain (\mathbb{R}) contains negative numbers and (0), which have no preimage. Step 3: If the range ((0,\infty)) differs from codomain (\mathbb{R}), the function is not onto.

Step 1: Put (y=\frac{x+1}{x-1}) and solve for (x). Step 2: (yx-y=x+1), so (x=\frac{y+1}{y-1}), defined for every (y\ne1). Step 3: If every codomain value gives a valid (x), the function is onto.

Step 1: In an onto function, every element of (B) must be used at least once. Step 2: Since (A) has (4) elements and (B) has (3), the possible count is (2,1,1). Step 3: If any codomain element has count (0), the function is not onto.

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 finite sets, subtract functions missing codomain elements.

Step 1: An onto function needs at least one preimage for every codomain element. Step 2: (A) has only (3) elements, so it cannot cover all (5) elements of (B). Step 3: For finite sets, if domain size is less than codomain size, onto functions are impossible.

Step 1: As (x\to\infty), (x^3-x\to\infty), and as (x\to-\infty), (x^3-x\to-\infty). Step 2: The polynomial is continuous, so it takes every real value in between. Step 3: For cubic polynomials, end behavior is very useful for onto checking.

Step 1: Since (x^4\ge0) and (x^2\ge0), (f(x)\ge0). Step 2: The codomain (\mathbb{R}) contains negative numbers, which are not in the range. Step 3: Even-power polynomials are often bounded below.

Step 1: The range of (\sin x) is ([-1,1]). Step 2: The codomain is also ([-1,1]), so every codomain value is obtained. Step 3: For trigonometric functions, compare the known range directly with the codomain.

Step 1: On ((-\frac{\pi}{2},\frac{\pi}{2})), the range of (\tan x) is all (\mathbb{R}). Step 2: For every real (y), (x=\tan^{-1}y) lies in this interval. Step 3: In (\tan x) questions, the domain is very important.

Step 1: (x^3+3x) is a continuous cubic polynomial. Step 2: As (x\to-\infty), it goes to (-\infty), and as (x\to\infty), it goes to (\infty). Step 3: Continuous cubic polynomials are often onto from (\mathbb{R}) to (\mathbb{R}).

Step 1: (|x-5|) has minimum value (0) and can take every non-negative value. Step 2: For any (y\ge0), choosing (x=5+y) gives (f(x)=y). Step 3: If range and codomain are both ([0,\infty)), the function is onto.

Step 1: (|x-5|) can never be negative. Step 2: The codomain (\mathbb{R}) contains (-2), but it has no preimage. Step 3: For absolute value functions from (\mathbb{R}) to (\mathbb{R}), check negative values.

Step 1: (2n+1) always gives odd integers. Step 2: The codomain (\mathbb{Z}) also has even integers, such as (2), with no preimage. Step 3: For integer functions, parity is a quick check.

Step 1: The codomain is the set of odd integers only. Step 2: Every odd (y) can be written as (y=2n+1), where (n=\frac{y-1}{2}\in\mathbb{Z}). Step 3: Changing the codomain can make the same formula onto.

Step 1: If (\mathbb{N}={1,2,3,\ldots}), then the minimum value of (n+2) is (3). Step 2: (1) and (2) are in the codomain but not in the range. Step 3: For natural numbers, always check the first few values.

Step 1: From (y=x^2+1), we get (x^2=y-1). Step 2: Since (y\ge1), (y-1\ge0), so (x=\sqrt{y-1}) is real. Step 3: To prove onto, solve for (x) in terms of (y) and check that (x) belongs to the domain.

Step 1: The domain is ([0,\infty)), and (x^2) can give every non-negative value. Step 2: For every (y\ge0), (x=\sqrt{y}) lies in the domain. Step 3: Whether (x^2) is onto depends on the codomain.

Step 1: Even when (x\ge0), (x^2\ge0). Step 2: The codomain (\mathbb{R}) contains values like (-1), which are not obtained. Step 3: Enlarging the codomain can make a function not onto.

Step 1: (\frac{x}{1+x^2}) is bounded and satisfies (-\frac{1}{2}\le f(x)\le\frac{1}{2}). Step 2: The codomain (\mathbb{R}) contains (2), but (f(x)=2) is impossible. Step 3: For rational expressions, identify range bounds.

Step 1: (f(x)=\frac{x}{1+x^2}) has maximum (\frac{1}{2}) at (x=1) and minimum (-\frac{1}{2}) at (x=-1). Step 2: All intermediate values are obtained by continuity. Step 3: If the codomain is a closed interval, check endpoints too.

Step 1: Take any codomain value (y\ne0). Step 2: For (f(x)=y), choose (x=\frac{1}{y}), which belongs to the domain because (y\ne0). Step 3: For reciprocal functions, excluding (0) from domain and codomain is important.