Onto function

Step 1: To prove onto, solve \(\tan x=y\). Step 2: For every real (y), \(x=\tan^{-1}y\) lies in the given interval. Step 3: Inverse trigonometric functions give a direct way to find preimages.

Step 1: The range of \(\tan^{-1}x\) is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). Step 2: Multiplying by \(\frac{2}{\pi}\) gives the range ((-1,1)), equal to the codomain. Step 3: Carefully apply the effect of a multiplier on the range.

Step 1: (\tan^{-1}x) never equals (\frac{\pi}{2}) or (-\frac{\pi}{2}). Step 2: After multiplying, (1) and (-1) are only limiting values, not actual images. Step 3: The difference between open and closed intervals can decide onto property.

Step 1: (x^3-x) is a continuous odd-degree polynomial and its range is (\mathbb{R}). Step 2: When the exponent takes all real values, (e^{x^3-x}) takes all positive values. Step 3: In composite functions, check the range of the inner function first.

Step 1: (h\circ g(x)=x^2+1). Step 2: Its range is ([1,\infty)), equal to the given codomain. Step 3: For composite functions, compare the final codomain with the actual range.

Step 1: Take any (c\in C). Step 2: Since (h) is onto, there is (b\in B) with (h(b)=c); since (g) is onto, there is (a\in A) with (g(a)=b). Step 3: Then ((h\circ g)(a)=c), so the composite function is onto.

Step 1: Every element of (C) is obtained as (h(g(a))). Step 2: This means every element of the codomain (C) is in the image of (h). Step 3: If a composite function is onto, the outer function must be onto.

Step 1: If (a\ne0), then for any (y\in\mathbb{R}), (x=\frac{y-b}{a}) is real. Step 2: If (a=0), the function becomes constant and cannot give all real values. Step 3: For a linear function, the coefficient of (x) must be non-zero.

Step 1: If (a>0), the range is ([1,\infty)). Step 2: If (a<0), the range is ((-\infty,1]), and if (a=0), only (1) is obtained. Step 3: In no case is the range all of (\mathbb{R}), so it is not onto.

Step 1: (x^5+x) is a continuous odd-degree polynomial. Step 2: It tends to (\infty) as (x\to\infty) and to (-\infty) as (x\to-\infty). Step 3: At expert level, combine continuity with end behavior to conclude.

Step 1: Since (x^6\ge0) and (x^2\ge0), (x^6+x^2+1\ge1). Step 2: The codomain (\mathbb{R}) contains (0), but it cannot be in the range. Step 3: Once a lower bound is found, look for codomain values below it.

Step 1: At (x=0), the value (1) is obtained, and large values are also obtained. Step 2: For (x^6+x^2+1=2), putting (t=x^2\ge0) gives (t^3+t=1), which has a non-negative solution. Step 3: Therefore the option saying (2) is not obtained is wrong, so this row needs a valid conclusion.

Step 1: For every (x), the denominator (1+x^2) is positive, so the function is continuous on all of (\mathbb{R}). Step 2: As (x\to\infty), the value tends to (\infty), and as (x\to-\infty), it tends to (-\infty). Step 3: Continuity together with unbounded behavior in both directions proves onto property.

Step 1: (x+e^x) is continuous. Step 2: As (x\to-\infty), the term (x) drives the value to (-\infty), and as (x\to\infty), the value goes to (\infty). Step 3: When a continuous function crosses all real values, it is onto.

Step 1: Since (x^2\ge0) and (e^x>0), we have (x^2+e^x>0). Step 2: The codomain (\mathbb{R}) contains (-1), but it cannot be an image of any (x). Step 3: Positivity or a lower bound can quickly disprove onto property.

Step 1: ((e^x-1)^2\ge0), and at (x=0), the value is (0). Step 2: For any (y\ge0), set (e^x-1=\sqrt{y}), giving (x=\ln(1+\sqrt{y})), which is real. Step 3: For exponential expressions, logarithms help construct preimages.

Step 1: (e^x>0), so (e^x+1>1). Step 2: Hence ((e^x+1)^2>1), so (0) and (1) are in the codomain but are not images. Step 3: Before squaring, check the range of the inner expression.

Step 1: This is a continuous polynomial and its leading term is (x^3). Step 2: For very large positive (x), the value tends to (\infty), and for very large negative (x), it tends to (-\infty). Step 3: Lower-degree terms do not change the end behavior.

Step 1: Since (x^2+1\ge1), (f(x)\ge1), and (1) is obtained at (x=0). Step 2: For any (y\ge1), (x=\sqrt{y^2-1}) is real and gives (f(x)=y). Step 3: For square-root functions, solve for (x) and check the domain.

Step 1: Since (x^2+1\ge1), (\sqrt{x^2+1}\ge1). Step 2: The codomain ([0,\infty)) contains (\frac{1}{2}), but it is not an image. Step 3: If the codomain is larger than the actual range, the function is not onto.

Step 1: For every (y\in[0,1]), (x=\sqrt{y}\in[-1,1]), so (y) is an image. Step 2: But (x) and (-x) have the same image, so the function is not one-one. Step 3: Prove onto and one-one properties separately.

Step 1: For every (y\in[0,1]), (x=\sqrt{y}\in[0,1]), so the function is onto. Step 2: On ([0,1]), (x^2) is increasing, so different (x) values give different images. Step 3: Restricting the domain can make the same formula one-one.

Step 1: (|x^2-1|) is always non-negative, and (0) is obtained at (x=1). Step 2: For any (y\ge0), taking (x=\sqrt{y+1}) gives (|x^2-1|=y). Step 3: In modulus questions, the positive branch can make preimage construction easier.

Step 1: Values of (\sin x) run from (-1) to (1). Step 2: Adding (1) changes the range to ([0,2]), equal to the codomain. Step 3: Apply vertical shifts clearly while finding the range.

Step 1: The range of (\cos x) is ([-1,1]). Step 2: Multiplying by (2) gives the range ([-2,2]), the same as the given codomain. Step 3: In trigonometric functions, a multiplier changes the range size but all intermediate values are still attained.