Onto function

Step 1: The range of (\frac{x}{1+|x|}) is ((-1,1)). Step 2: Multiplying by (2) changes the range to ((-2,2)). Step 3: Since the codomain is also ((-2,2)), the function is onto.

Step 1: The function value remains between (-2) and (2). Step 2: It is never equal to (-2) or (2), but both are included in the codomain. Step 3: In a closed interval codomain, endpoints must also be obtained.

Step 1: (|x+3|\ge0), so (f(x)\ge2). Step 2: For every (y\ge2), take (x=-3+(y-2)), then (f(x)=y). Step 3: The minimum value of a modulus function tells the start of its range.

Step 1: The minimum value of (|x+3|+2) is (2). Step 2: The codomain ([0,\infty)) contains (1), but the function cannot give (1). Step 3: If the codomain contains values below the range minimum, the function is not onto.

Step 1: (f'(x)=3x^2+2>0), so the function is increasing. Step 2: As (x\to\infty), (f(x)\to\infty), and as (x\to-\infty), (f(x)\to-\infty). Step 3: Such a continuous increasing function covers all of (\mathbb{R}).

Step 1: (x^4\ge0), so (x^4+2\ge2). Step 2: The codomain (\mathbb{R}) contains (1), which is not in the range. Step 3: Check the minimum value of even-power functions while testing onto nature.

Step 1: The range of (x^4+2) is ([2,\infty)). Step 2: For every (y\ge2), take (x=\sqrt[4]{y-2}), then (f(x)=y). Step 3: An even-power function can be onto a correctly shifted non-negative codomain.

Step 1: Since (x\ge1), (x-1\ge0), so the function is defined. Step 2: For every (y\ge0), take (x=y^2+1), then (f(x)=y). Step 3: For square-root functions, square the target value to find a preimage.

Step 1: A square-root value is always (0) or positive. Step 2: The codomain (\mathbb{R}) contains negative values such as (-1), which are not obtained. Step 3: A square-root function is not onto a full real codomain.

Step 1: Even numbers (2,4,6) give (0). Step 2: Odd numbers (1,3,5) give (1). Step 3: Both codomain values (0) and (1) are obtained, so the function is onto.

Step 1: All domain elements (2,4,6) are even. Step 2: Therefore, the function value is always (0), and (1) is never obtained. Step 3: If even one codomain value is missed, the function is not onto.

Step 1: (\sqrt{(x-4)^2}=|x-4|). Step 2: The range of (|x-4|) is ([0,\infty)), equal to the codomain. Step 3: Recognize (\sqrt{u^2}) as a modulus form.

Step 1: (\sqrt{(x-4)^2}=|x-4|). Step 2: Its value is never negative, while the codomain (\mathbb{R}) contains negative values. Step 3: Be careful when matching a modulus-type range with a real codomain.

Step 1: The cube root is defined for every real number. Step 2: For any (y\in\mathbb{R}), take (x=y^3+5), then (f(x)=y). Step 3: A cube-root function takes all real values on (\mathbb{R}).

Step 1: In the domain ([0,\infty)), (x) is non-negative. Step 2: Therefore (\sqrt[3]{x}) is also non-negative, and a value like (-1) is not obtained. Step 3: The range of a cube-root function depends on its domain.

Step 1: (\lfloor x+2\rfloor) gives integer values. Step 2: For any (k\in\mathbb{Z}), take (x=k-2), then (\lfloor x+2\rfloor=k). Step 3: A floor function can be onto an integer codomain.

Step 1: (\lfloor x+2\rfloor) is always an integer. Step 2: The codomain (\mathbb{R}) contains non-integer values such as (\frac{1}{3}), which are not in the range. Step 3: An integer-valued function is not onto the whole real codomain.

Step 1: (\lceil x-1\rceil) gives integer values. Step 2: For any (k\in\mathbb{Z}), take (x=k+1), then (\lceil x-1\rceil=k). Step 3: A ceiling function can cover an integer codomain completely.

Step 1: (\lceil x-1\rceil) always gives an integer. Step 2: The codomain (\mathbb{R}) contains a non-integer value like (\frac{5}{2}), but it is not obtained. Step 3: An integer-valued function should not be treated as onto a real codomain.

Step 1: Taking (x=0) gives (f(x)=0). Step 2: Taking (x=1) gives (f(x)=1). Step 3: Both codomain values are obtained, so the function is onto.

Step 1: The function gives only values (0) and (1). Step 2: The codomain also contains (2), but no (x) gives (f(x)=2). Step 3: A missed codomain value prevents onto nature.

Step 1: (|2x-6|) is always non-negative. Step 2: For every (y\ge0), set (2x-6=y), giving (x=\frac{y+6}{2}), and the function value becomes (y). Step 3: The modulus of a linear expression has range ([0,\infty)).

Step 1: Since ((x+4)^2\ge0), (f(x)\ge9). Step 2: For every (y\ge9), take (x=-4+\sqrt{y-9}), then (f(x)=y). Step 3: When the minimum value of a quadratic matches the start of the codomain, the function is onto.

Step 1: Put (f(x)=y), so (\frac{x}{3}=y). Step 2: This gives (x=3y), which is real for every real (y). Step 3: If every codomain value has a preimage, the function is onto.

Step 1: Since (|x-7|\ge0), (|x-7|-4\ge-4). Step 2: For every (y\ge-4), take (x=7+(y+4)), then (f(x)=y). Step 3: The minimum value of a modulus function helps identify its correct codomain.