Reflexive relation

Step 1: If (a\ne0), then for any (y\in\mathbb{R}), (x=\frac{y-b}{a}) is real. Step 2: Hence every (y) has a preimage. Step 3: A linear function on (\mathbb{R}) is onto when its slope is non-zero.

Step 1: If (k>0), values are at least (1). Step 2: If (k<0), values are at most (1), and if (k=0), the function is constant. Step 3: In no case does the range become all of (\mathbb{R}), so it cannot be onto.

Step 1: Since (x^2\ge0), (x^2+2\ge2). Step 2: For every (y\ge2), taking (x=\sqrt{y-2}) gives (f(x)=y). Step 3: For square functions, identify the minimum value to get the range.

Step 1: (\sqrt{x}) is always non-negative. Step 2: The codomain (\mathbb{R}) includes negative values, which are not attained. Step 3: Remember that the range of the square root function is ([0,\infty)).

Step 1: (\sqrt{x}) is increasing on ([0,\infty)), so it is one-one. Step 2: For every (y\ge0), choose (x=y^2), then (f(x)=y). Step 3: Thinking through the inverse helps prove onto quickly.

Step 1: (\frac{1}{x}) is never (0), and (0) is excluded from the domain. Step 2: For any (y\ne0), choosing (x=\frac{1}{y}) gives (f(x)=y). Step 3: For reciprocal functions, solve for (x) in terms of (y).

Step 1: Let (y=\frac{x}{x+1}). Step 2: From (y(x+1)=x), we get (x=\frac{y}{1-y}), defined for (y\ne1). Step 3: Since (1) is excluded from the codomain, every codomain value has a preimage.

Step 1: Setting (\frac{x}{x+1}=1) gives (x=x+1), which is impossible. Step 2: So (1) is not in the range, while the codomain is (\mathbb{R}). Step 3: In rational functions, identify the impossible output value.

Step 1: The function value is always between (0) and (1).
Step 2: For any (0<y<1), solving gives \(x=\ln\frac{y}{1-y}\).
Step 3: When the codomain is an open interval, endpoints need not be attained.

Step 1: The function values lie between (0) and (1). Step 2: Neither (0) nor (1) is attained for any real (x), but both are in the codomain. Step 3: The difference between open and closed intervals is crucial in onto questions.

Step 1: (x^2+2x=(x+1)^2-1). Step 2: The minimum value is (-1), and the range is ([-1,\infty)). Step 3: Completing the square is a clean method to test onto for quadratics.

Step 1: (f(x)=(x+1)^2-1) and (x\ge-1). Step 2: At (x=-1), the value is (-1), and as (x) increases all larger values occur. Step 3: Use the domain to select the correct branch of the parabola.

Step 1: If (x<0), then (|x|+x=-x+x=0). Step 2: If (x\ge0), then (|x|+x=2x), so values are non-negative. Step 3: Use piecewise behavior to find the range.

Step 1: The range of the function is ([0,\infty)). Step 2: For (y\ge0), choose (x=\frac{y}{2}), then (f(x)=y). Step 3: Use the branch that covers the whole codomain to build the preimage.

Step 1: The minimum value of (|x-3|) is (0). Step 2: For every (y\ge0), choosing (x=3+y) gives (|x-3|=y). Step 3: For modulus functions, distance interpretation makes the range easy.

Step 1: An absolute value is never negative. Step 2: The codomain (\mathbb{R}) contains values like (-1), which cannot be (f(x)). Step 3: A too-large codomain can make a function non-onto.

Step 1: (x^3) takes all real values. Step 2: (x^3+1) also takes all real values, because for any (y), (x=\sqrt[3]{y-1}) works. Step 3: A vertical shift of a cubic does not destroy onto behavior over (\mathbb{R}).

Step 1: (x^3+x) is defined and continuous for all real (x). Step 2: As (x\to\infty), the value goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Step 3: Continuity plus end behavior shows onto over (\mathbb{R}).

Step 1: (x^4\ge0) and (x^2\ge0). Step 2: Therefore (f(x)\ge0), so no negative real number is an image. Step 3: For sums of even powers, focus on the minimum value and range.

Step 1: The minimum value of (x^4+x^2) is (0). Step 2: Put (t=x^2\ge0); then (f=t^2+t), increasing continuously from (0) to (\infty). Step 3: Use a helper variable to find the range of difficult polynomials.