Onto function

Step 1: For onto nature, every value of the codomain must be obtained. Step 2: For any (y\in\mathbb{R}), take (x=y-5), then (f(x)=y). Step 3: A linear function (ax+b) with (a\ne0) is onto from (\mathbb{R}) to (\mathbb{R}).

Step 1: The value of (x^2) is always (0) or positive. Step 2: The codomain (\mathbb{R}) contains negative values such as (-1), but they are not in the range. Step 3: To test onto nature, compare range with codomain.

Step 1: The range of this function is ([0,\infty)). Step 2: The codomain is also ([0,\infty)), so every codomain value is obtained. Step 3: The same formula may become onto or not onto depending on the codomain.

Step 1: The codomain (B) has elements (a) and (b). Step 2: (a) has preimage (1), and (b) has preimage (2). Step 3: For finite sets, check each codomain element in the mapping list.

Step 1: For onto nature, every element of (B) must have a preimage. Step 2: The given pairs include (p) and (q), but not (r). Step 3: Identifying the missing codomain element is the easiest method.

Step 1: Put (f(x)=y), so (2x-7=y). Step 2: This gives (2x=y+7), hence (x=\frac{y+7}{2}). Step 3: If a real (x) exists for every real (y), the function is onto.

Step 1: For (n\in\mathbb{N}), the smallest value of (n+1) is (2). Step 2: The codomain (\mathbb{N}) contains (1), but no (n) satisfies (n+1=1). Step 3: For natural-number functions, check the smallest codomain values first.

Step 1: For any (y\in\mathbb{Z}), take (n=y+3). Step 2: (y+3) is also an integer, and (f(y+3)=y). Step 3: A shift function on integers is usually onto.

Step 1: (2n) is always an even integer. Step 2: The codomain (\mathbb{Z}) contains odd integers such as (1), which are not in the range. Step 3: While testing onto nature, look for types of values that are missing.

Step 1: The range of (e^x) is ((0,\infty)). Step 2: The codomain is also ((0,\infty)), so every codomain value is obtained. Step 3: Remembering the range of exponential functions helps decide onto nature quickly.

Step 1: (e^x) is always positive. Step 2: The codomain (\mathbb{R}) contains (0) and negative values, but they are not in the range. Step 3: If the codomain is larger than the range, the function is not onto.

Step 1: The range of (\ln x) is all real numbers. Step 2: For every (y\in\mathbb{R}), take (x=e^y>0), then (\ln x=y). Step 3: The logarithmic function is onto from its natural domain to (\mathbb{R}).

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: Remember the ranges of trigonometric functions.

Step 1: The value of (\sin x) always lies between (-1) and (1). Step 2: The codomain (\mathbb{R}) contains (2), but (\sin x=2) is impossible for real (x). Step 3: Be careful when trigonometric functions have (\mathbb{R}) as codomain.

Step 1: The range of (\cos x) is ([-1,1]). Step 2: The codomain is also ([-1,1]), so every codomain value is obtained. Step 3: If range and codomain are equal, the function is onto.

Step 1: For every real (y), (x=\sqrt[3]{y}) is real. Step 2: For this (x), (f(x)=x^3=y). Step 3: The cube function is onto from (\mathbb{R}) to (\mathbb{R}).

Step 1: Take any codomain value (y\ge0). Step 2: Put (x=y^2), then (x\ge0) and (\sqrt{x}=y). Step 3: The square-root function is onto its natural codomain.

Step 1: (|x|) is always (0) or positive. Step 2: The codomain (\mathbb{R}) contains negative values such as (-2), but they are not in the range. Step 3: First identify the range of a modulus function.

Step 1: The range of (|x|) is ([0,\infty)). Step 2: The codomain is also ([0,\infty)), so every codomain value is obtained. Step 3: With the correct codomain, a modulus function can be onto.

Step 1: In an onto function, every element of (B) needs at least one preimage. Step 2: (A) has only (3) elements while (B) has (5), so covering all elements of (B) is impossible. Step 3: For finite sets, onto requires (|A|\ge|B|).

Step 1: If a function is onto, every codomain element is reached by at least one domain element. Step 2: Therefore, the domain cannot have fewer elements than the codomain. Step 3: For finite sets, this is a necessary condition.

Step 1: (f(A)) is the range of the function. Step 2: Onto means the range is equal to the codomain. Step 3: Remember the definition of onto: (f(A)=B).

Step 1: A constant function gives only one value for every (x). Step 2: Here the range is only ({3}), while the codomain is (\mathbb{R}). Step 3: A constant function is not onto a codomain with more than one value.

Step 1: The codomain elements are (4,5,6). Step 2: All three are images of some domain elements. Step 3: For finite sets, if every codomain element appears as an image, the function is onto.

Step 1: The images listed are only (4) and (5). Step 2: The codomain element (6) is not the image of any (x). Step 3: A missing codomain element prevents onto nature.