Onto function

Step 1: For onto nature, every real value in the codomain must be obtained. Step 2: For any (y\in\mathbb{R}), take (x=y-12), then (f(x)=y). Step 3: A linear shift function on (\mathbb{R}) is onto.

Step 1: Put (f(x)=y), giving (-3x+6=y). Step 2: Then (-3x=y-6), so (x=\frac{6-y}{3}). Step 3: Since a real (x) exists for every real (y), the function is onto.

Step 1: ((x-1)^2\ge0), so (f(x)\ge4). Step 2: For every (y\ge4), take (x=1+\sqrt{y-4}), then (f(x)=y). Step 3: When the minimum value matches the codomain start, onto nature is easy to verify.

Step 1: The minimum value of ((x-2)^2+1) is (1). Step 2: The codomain ([0,\infty)) contains (0), but the function cannot produce (0). Step 3: While checking onto nature, always inspect the starting values of the codomain.

Step 1: ((x+1)^2\ge0), so (3-(x+1)^2\le3). Step 2: For every (y\le3), take (x=-1+\sqrt{3-y}), then (f(x)=y). Step 3: A downward quadratic can be onto its correct codomain.

Step 1: The maximum value of (4-x^2) is (4). Step 2: The codomain ((-\infty,5]) contains (5), but no real (x) can make (4-x^2=5). Step 3: If the codomain contains a value above the maximum range, the function is not onto.

Step 1: Take any (y\in\mathbb{Z}) from the codomain. Step 2: (n=y+8) is also an integer and (f(n)=y). Step 3: A fixed shift on integers covers all integers.

Step 1: (4n+1) always gives an integer leaving remainder (1) when divided by (4). Step 2: The codomain (\mathbb{Z}) contains (2), which is not of this form. Step 3: Remainder patterns are useful in integer-function onto questions.

Step 1: For (n\in\mathbb{N}), the smallest value of (n+5) is (6). Step 2: Hence (1,2,3,4,5) from the codomain are not in the range. Step 3: For natural numbers, quickly identify the missed initial values.

Step 1: The codomain is the natural numbers starting from (6). Step 2: For every (y\ge6), (n=y-5) is natural and (f(n)=y). Step 3: Choosing the codomain according to the range makes the function onto.

Step 1: The range of (e^x) is ((0,\infty)). Step 2: Adding (3) changes the range to ((3,\infty)), equal to the codomain. Step 3: Remember the open endpoint of an exponential range correctly.

Step 1: (e^x) is never (0). Step 2: Therefore (e^x+3) is always greater than (3), but the codomain contains (3). Step 3: The difference between open and closed intervals is important in onto questions.

Step 1: The range of (\ln x) is (\mathbb{R}). Step 2: Subtracting (2) still allows all real values. Step 3: A vertical shift of a logarithmic function does not change its onto nature over (\mathbb{R}).

Step 1: Since (x>1), (x-1>0). Step 2: For any (y\in\mathbb{R}), take (x=1+e^y), then (\ln(x-1)=y). Step 3: A logarithm can give all real values when its inside quantity stays positive.

Step 1: On ([0,2\pi]), (\sin x) takes all values from (-1) to (1). Step 2: The codomain is ([-1,1]), so every codomain value is obtained. Step 3: On a full period, identifying trigonometric range is easy.

Step 1: On ([0,\frac{\pi}{6}]), the range of (\sin x) is ([0,\frac{1}{2}]). Step 2: The codomain ([0,1]) contains (\frac{3}{4}), which is not in the range. Step 3: On a small trigonometric interval, the range can also be small.

Step 1: On ([0,\frac{\pi}{3}]), the range of (\cos x) is ([\frac{1}{2},1]). Step 2: The codomain ([-1,1]) contains negative values, which are not obtained. Step 3: When the domain is restricted, the range must be checked again.

Step 1: On this interval, (\tan x) takes all real values. Step 2: For every (y\in\mathbb{R}), there is an (x) in this interval such that (\tan x=y). Step 3: Remember the standard range of (\tan x) as (\mathbb{R}).

Step 1: On ([0,\frac{\pi}{4}]), the range of (\tan x) is ([0,1]). Step 2: The codomain ([0,2]) contains (2), but it is not in the range. Step 3: If the codomain is larger than the range, the function is not onto.

Step 1: The codomain (B) has elements (a,b,c). Step 2: The images include (a), (b), and (c). Step 3: For finite sets, if every codomain element appears as an image, the function is onto.

Step 1: The listed images are only (p) and (q). Step 2: The codomain element (r) is not obtained from any domain element. Step 3: Even one missed codomain element destroys onto nature.

Step 1: In an onto function, every element of (B) needs a preimage. Step 2: (A) has only (4) elements while (B) has (6), so covering all elements is impossible. Step 3: If (|A|<|B|), the number of onto functions is (0).

Step 1: The total number of functions is (2^4=16). Step 2: The two non-onto functions send all elements only to (0) or only to (1). Step 3: Hence the number of onto functions is (16-2=14).

Step 1: At (x=0), the value (0) is obtained. Step 2: For (0\le y<1), solving (\frac{x^2}{1+x^2}=y) gives (x^2=\frac{y}{1-y}). Step 3: Hence every value in ([0,1)) is obtained.

Step 1: In (\frac{x^2}{1+x^2}), the denominator is always larger than the numerator for real (x). Step 2: Therefore the value stays below (1), so (1) is never obtained. Step 3: If an endpoint of a closed codomain is missed, the function is not onto.