Onto function

Step 1: This is an odd-degree polynomial with positive leading coefficient. Step 2: As (x\to\infty), the value goes to (\infty), and as (x\to-\infty), it goes to (-\infty), so every real target value is obtained. Step 3: In hard questions, first check the end behavior of odd-degree polynomials.

Step 1: We can write (x^4-4x^2=(x^2-2)^2-4). Step 2: Its minimum value is (-4), so values like (-5) are not obtained. Step 3: For even-degree polynomials, completing the square helps find the range.

Step 1: (f(x)=(x^2-2)^2-4). Step 2: Its range is ([-4,\infty)), exactly the given codomain. Step 3: For onto, the equality of range and codomain is more important than the formula alone.

Step 1: (x+\sin x) is continuous. Step 2: As (x\to\infty), it goes to (\infty), and as (x\to-\infty), it goes to (-\infty), because the (x)-term dominates. Step 3: When a bounded trigonometric term is added to (x), check end behavior.

Step 1: (\cos x) is bounded, while (x) grows without bound in both directions. Step 2: Hence (x-\cos x) also goes without bound in both directions and is continuous. Step 3: Continuity plus end behavior can prove onto.

Step 1: As (x\to0^+), (x-\frac{1}{x}\to-\infty). Step 2: As (x\to\infty), (x-\frac{1}{x}\to\infty), and the function is continuous. Step 3: For open-domain questions, use limits at the ends to understand the range.

Step 1: For (x>0), (x+\frac{1}{x}\ge 2). Step 2: The codomain (\mathbb{R}) contains (1) and negative numbers, which are not obtained. Step 3: Identifying the minimum of (x+\frac{1}{x}) is very useful in exams.

Step 1: The minimum value of (x+\frac{1}{x}) is (2), attained at (x=1). Step 2: As (x\to0^+) or (x\to\infty), the value goes to (\infty), so all of ([2,\infty)) is obtained. Step 3: After finding the minimum, compare it with the codomain.

Step 1: Put (y=\frac{2x+3}{x-1}), so (yx-y=2x+3). Step 2: From ((y-2)x=y+3), we get (x=\frac{y+3}{y-2}), valid for (y\ne2). Step 3: For fractional functions, solving (x) in terms of (y) is a strong method.

Step 1: If (\frac{x+1}{x+2}=1), then (x+1=x+2), which is impossible. Step 2: Thus (1) is not in the range, while the codomain is (\mathbb{R}). Step 3: In linear fractional functions, the horizontal limiting value often indicates the missing value.

Step 1: From (y=\frac{x+1}{x+2}), we get (yx+2y=x+1). Step 2: Then ((y-1)x=1-2y), so (x=\frac{1-2y}{y-1}), valid for (y\ne1). Step 3: Removing the missing value from the codomain can make the function onto.

In option B, every \(y\geq0\) in the codomain has a preimage: take \(x=\sqrt{y}\) or \(x=-\sqrt{y}\), so \(x^2=y\). Options C and D never produce 0. Exam tip: compare the range with the stated codomain.

Step 1: Put \(t=x^2\ge0\), so the function becomes \(\frac{t+1}{t+2}\). Step 2: At \(t=0\), the value is \(\frac{1}{2}\), and as \(t\to\infty\), it approaches \(1\) without reaching it. Step 3: Replacing \(x^2\) by \(t\) simplifies hard range questions.

Step 1: For onto, (ax+b=y) must be solvable for every (y\in\mathbb{R}). Step 2: If (a\ne0), then (x=\frac{y-b}{a}) is real for every (y). Step 3: In a linear function, (a=0) makes it constant.

Step 1: If (a=0), the function is constant. Step 2: If (a>0), the range is ([1,\infty)), and if (a<0), the range is ((-\infty,1]). Step 3: None of these ranges equals all of (\mathbb{R}).

Step 1: (x^3+a x) is an odd-degree polynomial. Step 2: Since the leading term is (x^3), the value goes to (\infty) as (x\to\infty) and to (-\infty) as (x\to-\infty). Step 3: A continuous odd-degree polynomial from (\mathbb{R}) to (\mathbb{R}) is onto.

Step 1: (x^3-x) is an odd-degree polynomial and has range (\mathbb{R}). Step 2: For (e^u) with (u\in\mathbb{R}), the range is ((0,\infty)). Step 3: In composite functions, first check the range of the inner function.

Step 1: Since (x^2\ge0), (e^{x^2}\ge1). Step 2: The codomain (\mathbb{R}) contains (0) and negative values that are not obtained. Step 3: For exponential functions, the minimum of the exponent decides the range.

Step 1: At (x=0), (e^{x^2}=1). Step 2: For any (y\ge1), we can take (x=\sqrt{\ln y}). Step 3: Prove onto clearly by constructing (x) from the target value.

Step 1: For any (y\in\mathbb{R}), (x=e^y) is positive. Step 2: Then (\ln x=\ln(e^y)=y). Step 3: The relation between logarithm and exponential helps prove onto quickly.

Step 1: The codomain is only ((0,\infty)), so only positive target values must be checked. Step 2: For every (y>0), (x=e^y>0) and (\ln x=y). Step 3: For onto, values outside the codomain do not matter.

Step 1: This function always gives a value between (0) and (1). Step 2: For every \(y\in(0,1)\), taking \(x=\ln\left(\frac{y}{1-y}\right)\) gives \(f(x)=y\). Step 3: In hard onto questions, solving for (x) from target (y) is very useful.

Step 1: (\frac{1}{1+e^{-x}}) is always greater than (0) and less than (1). Step 2: The codomain ([0,1]) includes (0) and (1), but they are never attained. Step 3: The difference between open and closed intervals can decide onto.

Step 1: Total functions are (2^5=32). Step 2: Non-onto functions send all (5) elements to just one target; there are (2) such functions. Step 3: Therefore onto functions are (32-2=30).

Step 1: Total functions are (3^5=243). Step 2: The number missing at least one target is (3\cdot2^5-3\cdot1^5=96-3=93). Step 3: Onto functions are (243-93=150).