Onto function

Step 1: (\tan x) takes all real values on this interval. Step 2: For any \(y\in\mathbb{R}\), \(x=\tan^{-1}y\) lies in the given interval. Step 3: Using an inverse expression is an effective way to prove onto.

Step 1: (\tan x) is undefined at (x=\frac{\pi}{2}+k\pi). Step 2: These values are in (\mathbb{R}), so (f:\mathbb{R}\to\mathbb{R}) is not well-defined. Step 3: A function must be well-defined before onto nature can be tested.

Step 1: The value of this function always lies between (0) and (1). Step 2: For every (y\in(0,1)), taking (x=\ln\frac{y}{1-y}) gives (f(x)=y). Step 3: Solving for (x) from a target value is a strong proof of onto.

Step 1: For every real (x), (0<\frac{1}{1+e^{-x}}<1). Step 2: The codomain ([0,1]) contains (0) and (1), but they are never obtained. Step 3: A value approached as a limit need not belong to the range.

Step 1: (\sqrt{x}) is always non-negative for (x\ge0). Step 2: For every (y\in[0,\infty)), take (x=y^2), then (\sqrt{x}=y). Step 3: For root functions, find (x) in terms of the target value.

Step 1: (\sqrt{x}\ge0) for all (x\ge0). Step 2: The codomain (\mathbb{R}) contains negative numbers such as (-2), which are not obtained. Step 3: If the range is smaller than the codomain, the function is not onto.

Step 1: (\sqrt[3]{x}) is defined for all real (x). Step 2: For every real (y), take (x=y^3), then (f(x)=y). Step 3: The cube-root function can give both negative and positive values.

Step 1: An absolute value is always non-negative. Step 2: The codomain (\mathbb{R}) contains negative values, which cannot be obtained from (f(x)). Step 3: For absolute value functions, the range usually starts from ([0,\infty)).

Step 1: The smallest value of (|x-2|) is (0), obtained at (x=2). Step 2: For every (y\ge0), taking (x=2+y) gives (|x-2|=y). Step 3: In absolute value functions, use (x=a+y) or (x=a-y) to obtain a target value.

Step 1: (x^3+x) is continuous on all real numbers. Step 2: For very large positive (x), values become very large positive, and for very large negative (x), values become very large negative. Step 3: Continuity with unbounded behavior on both sides ensures every real value is obtained.

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

Step 1: Put (t=x^2\ge0), so (x^4-x^2=t^2-t). Step 2: The minimum value of (t^2-t) is (-\frac{1}{4}), so (-1) cannot be obtained. Step 3: For fourth-degree expressions in (x^2), substituting (t=x^2) helps find the range.

The function \(f(x)=x^3-x\) is a continuous cubic polynomial. As \(x\to\infty\), \(f(x)\to\infty\), and as \(x\to-\infty\), \(f(x)\to-\infty\). By continuity, it takes every real value, so its range is \(\mathbb{R}\) and it is onto. In contrast, \(e^x\) takes only positive values and is therefore not onto \(\mathbb{R}\). Exam tip: to test onto-ness, check whether every element of the stated codomain has at least one preimage.

Step 1: For (x\in[0,1]), (x^2\in[0,1]). Step 2: For every (y\in[0,1]), (x=\sqrt{y}) also lies in ([0,1]) and (x^2=y). Step 3: In closed interval problems, the preimage must belong to the same domain.

Step 1: For (x\in[0,1]), (0\le x^2\le1). Step 2: The codomain ([0,2]) contains values like (1.5), which are not obtained. Step 3: When the domain is restricted, maximum and minimum values quickly give the range.

Step 1: (\lfloor x\rfloor) is always an integer. Step 2: The codomain (\mathbb{R}) contains non-integer values like (\frac{1}{2}), which are not obtained. Step 3: For the greatest integer function, identify the range first.

Step 1: (\lfloor x\rfloor) is always an integer. Step 2: For every (n\in\mathbb{Z}), taking (x=n) gives (\lfloor x\rfloor=n). Step 3: With codomain (\mathbb{Z}), the greatest integer function becomes onto.

Step 1: (x+\sin x) is continuous on all real numbers. Step 2: (\sin x) is bounded, while (x) grows without bound in both directions, so (f(x)) also goes to both infinities. Step 3: A bounded addition often does not destroy onto behavior of a linear term.

Step 1: (x-\cos x) is continuous. Step 2: The (x) part goes to (\pm\infty), while (\cos x) stays in ([-1,1]), so the total expression crosses every real level. Step 3: With a bounded trigonometric part and a linear part, check limits.

Step 1: (\ln x) is defined only for (x>0). Step 2: For every (y\in\mathbb{R}), take (x=e^y>0), then (\ln x=y). Step 3: Logarithmic and exponential functions are useful as inverse pairs.

Step 1: For (x\in(0,1)), (\ln x<0). Step 2: But the codomain ((0,\infty)) contains only positive values, so not all outputs lie in the codomain. Step 3: With a wrong codomain, well-definedness fails before onto checking.

Step 1: (x^2\ge0), so (x^2+1\ge1). Step 2: For every (y\ge1), take (x=\sqrt{y-1}), then (f(x)=y). Step 3: The range of (x^2+a) is directly ([a,\infty)).

Step 1: At (x=0), (f(0)=1). Step 2: But (1) is not in the codomain ((1,\infty)), so the function is not well-defined for this codomain. Step 3: First match every output with the given codomain.

Step 1: Let any codomain value be (y\ne0). Step 2: Taking (x=\frac{1}{y}) gives (x\ne0) and (f(x)=y). Step 3: For reciprocal functions, write (x) in terms of (y) to test onto.

Step 1: Take any target value (y\in[-1,1)) and write (y=\frac{x^2-1}{x^2+1}). Step 2: Solving gives (x^2=\frac{1+y}{1-y}), which is non-negative because (y\ge -1) and (y<1), so a real (x) exists. Step 3: To prove onto, finding a valid preimage for every target value is the strongest method.