Onto function

Step 1: This is a continuous odd-degree polynomial. Step 2: As (x\to\infty), the value goes to (\infty), and as (x\to-\infty), it goes to (-\infty), so every real value is attained. Step 3: Do not be confused by turning points; end behavior is very useful for onto property.

Step 1: Since (x^4\ge0) and (x^2\ge0), we have (x^4+x^2\ge0). Step 2: The codomain (\mathbb{R}) contains (-1), but no real (x) can map to it. Step 3: For sums of even powers, check negative codomain values first.

Step 1: (x^2-6x+11=(x-3)^2+2). Step 2: Its range is ([2,\infty)), so the codomain must be ([2,\infty)) for onto property. Step 3: Complete the square to find the minimum value of a quadratic.

Step 1: (f(x)=(x-1)^2+4), and (1\in[-2,3]). Step 2: The minimum is (4), and at the endpoints (f(-2)=13), (f(3)=8), so the maximum is (13). Step 3: On a closed interval, check the vertex and both endpoints.

Step 1: Put (y=\frac{3x+1}{x-2}), then solving gives (x=\frac{2y+1}{y-3}). Step 2: Since (y\ne3) in the codomain, this (x) is defined and it also satisfies (x\ne2). Step 3: In rational functions, relate the missed value to the denominator restriction.

Step 1: At (x=0), (0) is obtained, and at (x=1), (\frac{1}{2}) is obtained. Step 2: (\frac{x^2}{1+x^2}<1) for every real (x), so (1) is never obtained. Step 3: When the denominator is larger than the numerator, check endpoint limit values carefully.

Step 1: For every (x), (e^x>0), so the value lies between (0) and (1). Step 2: For any (0<y<1), (e^x=\frac{1-y}{y}), so (x=\ln\frac{1-y}{y}) is real. Step 3: In an open interval codomain, endpoints are not required.

Step 1: (e^x+e^{-x}\ge2), so (f(x)\ge0). Step 2: At (x=0), (0) is obtained, and as (|x|) increases, the value continuously goes to (\infty), so every non-negative value is obtained. Step 3: Use both inequality and continuity in such questions.

Step 1: (x+\sin x) is continuous. Step 2: (\sin x) is bounded, but (x) is unbounded, so the function goes unbounded in both directions. Step 3: A bounded added term does not stop the unbounded linear behavior.

Step 1: Since (x^2\ge0) and (\sin x\ge-1), (f(x)\ge-1). Step 2: The codomain (\mathbb{R}) contains values like (-2), but they cannot be images. Step 3: Even without the exact range, a lower bound can disprove onto property.

Step 1: The total number of functions is (3^5). Step 2: Subtract functions missing one codomain element, (\binom{3}{1}2^5), and add those missing two, (\binom{3}{2}1^5). Step 3: Use inclusion-exclusion for counting onto functions.

Step 1: For finite sets of equal size, an onto function is automatically one-one. Step 2: So it is a permutation of (4) elements, giving (4!) functions. Step 3: When domain and codomain have equal size, onto and one-one go together.

Step 1: In an onto function, every codomain element must have at least one preimage. Step 2: The domain has only (3) elements and the codomain has (5), so covering all (5) elements is impossible. Step 3: If the domain size is less than the codomain size, no onto function exists.

Step 1: (2n+1) is always an odd integer. Step 2: The codomain (\mathbb{Z}) contains even integers such as (0), (2), and (-2), but they are not images. Step 3: For integer functions, parity is a quick way to test onto property.

Step 1: Take any odd integer (m) in the codomain. Step 2: It can be written as (m=2k+1) for some integer (k), so (f(k)=m). Step 3: A function becomes onto when the codomain matches its actual range.

\(h(1)=h(-2)\) only proves that the function is not one-one. As \(x\to\infty\), \(h(x)\to\infty\), and as \(x\to-\infty\), \(h(x)\to-\infty\); hence every real value is attained. Exam tip: repeated outputs test one-one-ness, not onto-ness.

Step 1: (x^2+2x=(x+1)^2-1). Step 2: Its range is ([-1,\infty)), exactly equal to the codomain. Step 3: Writing in completed-square form makes onto property clear.

Step 1: Take any (y\ge0) from the codomain. Step 2: Choosing (x=\sqrt{y+1}) gives (x\ge1) and (f(x)=y). Step 3: In restricted domains, always check that the found (x) really lies in the domain.

Step 1: On ((0,\infty)), (x-\frac{1}{x}) is continuous. Step 2: As (x\to0^+), it tends to (-\infty), and as (x\to\infty), it tends to (\infty). Step 3: A continuous function crossing all real values gives onto property.

Step 1: (x+\frac{1}{x}\ge2) for (x>0). Step 2: The codomain ((0,\infty)) contains (1), but it is not an image. Step 3: The standard inequality (x+\frac{1}{x}\ge2) is useful in such problems.

Step 1: (x+\frac{1}{x}\ge2), and (2) is obtained at (x=1). Step 2: For every (y>2), the equation (x+\frac{1}{x}=y) has a positive solution. Step 3: Check separately whether the minimum value is actually attained or only approached.

Step 1: If (x\in[n,n+1)), then (\lfloor x\rfloor=n) and (f(x)\in[2n,2n+1)). Step 2: Hence values in intervals such as ((1,2)) are missed. Step 3: For floor functions, split the domain into unit intervals.

Step 1: Take any integer (k) from the codomain. Step 2: Putting (x=k) gives (\lceil x\rceil=k), so every integer is an image. Step 3: For integer-valued functions with codomain (\mathbb{Z}), directly find a preimage.

Step 1: On ([0,\pi]), (\cos x) decreases continuously from (1) to (-1). Step 2: Hence every value in ([-1,1]) is attained. Step 3: For trigonometric functions, check the range on the selected interval.

Step 1: On \(\left[0,\frac{\pi}{2}\right]\), the range of (\sin x) is ([0,1]). Step 2: The codomain ([-1,1]) also contains negative values, which are not images. Step 3: Along with the standard range, find the range on the given domain.