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+9), then (f(x)=y). Step 3: A non-zero slope linear function from (\mathbb{R}) to (\mathbb{R}) is onto.

Step 1: Put (4x+1=y). Step 2: This gives (4x=y-1), so (x=\frac{y-1}{4}). Step 3: If such a real (x) exists for every real (y), the function is onto.

Step 1: (x^2\ge0), so (x^2+5\ge5). Step 2: For every (y\ge5), take (x=\sqrt{y-5}), then (f(x)=y). Step 3: The range is ([5,\infty)), equal to the codomain, so the function is onto.

Step 1: The minimum value of (x^2+5) is (5). Step 2: The codomain (\mathbb{R}) contains (3), but (x^2+5=3) has no real solution. Step 3: Values missing from the range prevent onto nature.

Step 1: (x^2\ge0), so (7-x^2\le7). Step 2: For every (y\le7), take (x=\sqrt{7-y}), then (f(x)=y). Step 3: A downward-opening quadratic can be onto its correct codomain.

Step 1: The maximum value of (7-x^2) is (7). Step 2: The codomain (\mathbb{R}) contains (8), but (7-x^2=8) has no real solution. Step 3: Codomain values above the maximum are not in the range.

Step 1: Take any (y\in\mathbb{Z}) from the codomain. Step 2: (n=y-4) is also an integer and (f(n)=y). Step 3: A shift function on integers can obtain every integer value.

Step 1: (3n) is always a multiple of (3). Step 2: The codomain (\mathbb{Z}) contains (1), which cannot be written as (3n). Step 3: For multiple-based functions, look for integers that are missed.

Step 1: For (n\in\mathbb{N}), the smallest value of (n+3) is (4). Step 2: So (1,2,3) from the codomain are not in the range. Step 3: For natural-number functions, checking initial values is very useful.

Step 1: The codomain is the set of natural numbers starting from (4). Step 2: For every (y\ge4), (n=y-3) is a natural number 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: Therefore, the range of (e^x+2) is ((2,\infty)), equal to the codomain. Step 3: A vertical shift of an exponential function shifts its range upward.

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

Step 1: The range of (\ln x) is (\mathbb{R}). Step 2: (\ln x+4) also takes all real values, because shifting (\mathbb{R}) still gives (\mathbb{R}). Step 3: A logarithmic function on its natural domain covers all real values.

Step 1: For (0<x<1), (\ln x<0). Step 2: The codomain is ((0,\infty)), which does not contain negative values, so the given mapping is not properly a function into this codomain. Step 3: Before testing onto nature, check whether the function is well-defined.

Step 1: As (x) goes from (0) to (\pi), (\cos x) takes every value from (1) to (-1). Step 2: The codomain is ([-1,1]), so every value in it is obtained. Step 3: Identify the range carefully on restricted trigonometric intervals.

Step 1: On ([0,\frac{\pi}{2}]), the range of (\cos x) is ([0,1]). Step 2: The codomain ([-1,1]) contains negative values, which are not in the range. Step 3: Restricting the domain can change the range.

Step 1: On this interval, (\sin x) takes all values from (-1) to (1). Step 2: The codomain is ([-1,1]), so no codomain value is missed. Step 3: Remember the ranges on standard trigonometric intervals.

Step 1: On ([0,\pi]), the range of (\sin x) is ([0,1]). Step 2: The codomain ([-1,1]) contains negative values, but they are not obtained from this domain. Step 3: For trigonometric functions, changing the domain changes onto nature.

Step 1: On ((-\frac{\pi}{2},\frac{\pi}{2})), the range of (\tan x) is (\mathbb{R}). Step 2: Thus every real codomain value is obtained for some (x). Step 3: Remember the standard onto interval for (\tan x).

Step 1: On ([0,\frac{\pi}{4}]), (\tan x) takes values from (0) to (1). Step 2: The codomain (\mathbb{R}) contains values such as (2), which are not in the range. Step 3: A small range cannot cover a larger codomain.

Step 1: The codomain (B) has elements (a,b,c). Step 2: All three elements appear as images of some domain element. Step 3: For finite sets, look for every codomain element among the images.

Step 1: The listed images contain only (a) and (b). Step 2: The codomain element (c) is not obtained from any domain element. Step 3: If even one codomain element is missed, the function is not onto.

Step 1: For onto functions between finite sets, the domain should not have fewer elements than the codomain. Step 2: Here (5\ge2), so it is possible to cover all codomain elements. Step 3: When possibility is asked, first compare the sizes.

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

Step 1: The total number of functions is (2^3=8). Step 2: The non-onto functions are those sending all elements only to (p) or only to (q), so (2) functions. Step 3: Hence the number of onto functions is (8-2=6).