Onto function

Step 1: (\frac{1}{x}) can never be (0). Step 2: The codomain (\mathbb{R}) contains (0), so one codomain element is missing. Step 3: If a reciprocal has constant numerator, check whether (0) is obtained.

Step 1: Since (1+x^2\ge1), (0<f(x)\le1). Step 2: For any (y\in(0,1]), (x^2=\frac{1}{y}-1\ge0), so a real (x) exists. Step 3: To find range, rewrite the equation in terms of (x^2).

Step 1: (f(x)) is always greater than (0). Step 2: The codomain ([0,1]) contains (0), but no real (x) gives it. Step 3: Pay close attention to open and closed endpoints.

Step 1: On ([-1,1]), the range of (x^2) is ([0,1]). Step 2: For every (y\in[0,1]), (x=\sqrt{y}) lies in the domain. Step 3: Even on a restricted domain, a function can be onto if codomain is chosen correctly.

Step 1: On ([-1,1]), values of (x^2) lie in ([0,1]). Step 2: The codomain ([-1,1]) also contains negative values such as (-\frac{1}{2}), which are not obtained. Step 3: Having a similar-looking domain does not guarantee onto.

Step 1: Put (y=ax+b) and solve for (x). Step 2: (x=\frac{y-b}{a}) is real for every real (y) because (a\ne0). Step 3: A non-zero slope linear function from (\mathbb{R}) to (\mathbb{R}) is onto.

Step 1: The range of a constant function is only ({5}). Step 2: The codomain (\mathbb{R}) has values such as (6), (0), and (-1), which are not obtained. Step 3: A constant function is not onto over a large codomain.

Step 1: The values of the function are (p,q,p). Step 2: The codomain element (r) is not the image of any domain element. Step 3: In table-based questions, mark the missing codomain element.

Step 1: All three codomain elements (a,b,c) appear as images. Step 2: Repetition of (a) does not stop onto property. Step 3: Onto requires every codomain element at least once, not exactly once.

Step 1: In an onto function, every element of (B) needs at least one element from (A). Step 2: Therefore, (A) cannot have fewer elements than (B). Step 3: For finite sets, (n(A)\ge n(B)) is a necessary condition for onto.

Step 1: Take any element (c\in C). Step 2: Since (g) is onto, some (b\in B) has (g(b)=c); since (f) is onto, some (a\in A) has (f(a)=b). Step 3: In composition, trace the preimage backward.

Step 1: From (y=x^3+1), we get (x^3=y-1). Step 2: For every real (y), (\sqrt[3]{y-1}) is real. Step 3: Cube root is defined for every real number, so it helps prove onto.

Step 1: Since (x\ge0), (x^3+1\ge1). Step 2: The codomain (\mathbb{R}) contains (0) and negative values, which are not obtained. Step 3: Changing the domain restriction changes the range.

Step 1: From (y=x^3+1), (x=\sqrt[3]{y-1}). Step 2: If (y\ge1), then (y-1\ge0), so (x\ge0) and belongs to the domain. Step 3: Finding a valid preimage in the domain is the key condition for onto.

Step 1: Since (x^2\ge0), (x^2-3\ge-3). Step 2: For every (y\ge-3), choosing (x=\sqrt{y+3}) gives (f(x)=y). Step 3: Use the minimum value of a square function to decide its range.

Step 1: Since (x^2\ge0), (4-x^2\le4). Step 2: For every (y\le4), (x=\sqrt{4-y}) is real and gives (f(x)=y). Step 3: For a downward quadratic, use the maximum value to identify the range.

Step 1: Since (x^6\ge0), (x^6+1\ge1). Step 2: The codomain (\mathbb{R}) contains (0) and negative values, which are not obtained. Step 3: Even powers with a positive shift have a range bounded below.

Step 1: Since (x\ge1), (x^2-1\ge0). Step 2: For every (y\ge0), (x=\sqrt{y+1}\ge1), so it lies in the domain. Step 3: Always check the domain condition for the (x) you find.

Step 1: The range of (\ln x) is all (\mathbb{R}) for (x>0). Step 2: For any (y\in\mathbb{R}), (x=e^y>0), so (x) is in the domain. Step 3: Logarithmic and exponential functions help find preimages of each other.

Step 1: (\ln x) takes negative, zero, and positive real values. Step 2: Although every positive (y) has (x=e^y), many outputs of (\ln x) are outside ((0,\infty)). Step 3: A function (A\to B) must have every output inside (B), so with this codomain it is not a valid function into (B).

Step 1: (\lfloor x\rfloor) always gives an integer value. Step 2: The codomain (\mathbb{R}) contains non-integers such as (\frac{1}{2}), with no preimage. Step 3: The greatest integer function has range (\mathbb{Z}), not (\mathbb{R}).

Step 1: Put (y=\frac{3x+1}{x+2}) and solve for (x). Step 2: (yx+2y=3x+1), so (x=\frac{1-2y}{y-3}), which is defined for every (y\ne3). Step 3: In rational functions, the value that makes solving impossible is often excluded from the codomain.

Step 1: (x^2-2x+5=(x-1)^2+4), so its minimum value is (4). Step 2: The codomain (\mathbb{R}) contains (3), but the range is ([4,\infty)). Step 3: Completing the square is a fast and reliable exam method for finding range.

Step 1: On the given interval, (\sin x) increases continuously. Step 2: At (x=-\frac{\pi}{2}), the value is (-1), and at (x=\frac{\pi}{2}), the value is (1), so all values between them are obtained. Step 3: On a closed interval, checking endpoint values helps identify onto behavior quickly.

Step 1: Total functions are (3^6). Step 2: By inclusion-exclusion, onto functions (=3^6-\binom{3}{1}2^6+\binom{3}{2}1^6=729-192+3=540). Step 3: For counting onto functions, start with all functions and subtract those missing codomain elements.