Functions

Step 1: The expression (x^3+x) increases continuously as (x) increases, so two different inputs cannot give the same value. Step 2: Its values cover all real numbers from very negative to very positive. Step 3: A function that is both one-one and onto is also invertible.

Step 1: (f(1)=2) and (f(-1)=2), so the function is not one-one. Step 2: Since (x^2+1\geq 1), real values like (0) are not in the range. Step 3: For square functions, always check both domain and codomain before deciding the type.

Step 1: On both negative and positive sides, the function increases and the order remains consistent at (0). Step 2: Its values always lie between (-1) and (1), and every value in that interval is attained. Step 3: When the codomain is exactly the natural range, check for onto carefully.

Step 1: Adding (1) to different natural numbers gives different values, so the function is one-one. Step 2: (1) is in the codomain, but no (n\in\mathbb{N}) satisfies (n+1=1). Step 3: In natural number functions, the first element often helps test onto.

Step 1: (2n+1) gives different odd integers for different integers, so the function is one-one. Step 2: Even integers such as (0) are in the codomain but are not obtained from any integer (n). Step 3: For linear functions, the codomain can change whether the function is onto.

Step 1: (e^x) is strictly increasing, so different inputs give different outputs. Step 2: It is never (0) or negative, and every positive value can occur. Step 3: Because the codomain is ((0,\infty)), the function is onto.

Step 1: ((g\circ f)(x)) means (g(f(x))). Step 2: (g(2x-3)=(2x-3)^2+1=4x^2-12x+10). Step 3: Do not reverse the order in composition because (g\circ f) and (f\circ g) are usually different.

Step 1: Write (y=3x+4) and solve for (x) in terms of (y). Step 2: (y-4=3x), so (x=\frac{y-4}{3}). Step 3: Replace (y) by (x) to get (f^{-1}(x)=\frac{x-4}{3}).

Step 1: Let (y=\frac{x-2}{x+3}). Step 2: From (yx+3y=x-2), we get (x(y-1)=-(2+3y)). Step 3: For (y=1), the equation becomes impossible, so (1) is not in the range.

Step 1: If (f(a)=f(b)), then (g(f(a))=g(f(b))). Step 2: Since (g\circ f) is one-one, this forces (a=b). Step 3: Therefore (f) must be one-one, but (g) need not always be one-one.

Step 1: If (g\circ f) is onto, every element of the final codomain is obtained as (g(f(x))). Step 2: Hence every such element is also obtained as a value of (g). Step 3: Therefore (g) must be onto, but (f) need not be onto.

Step 1: In a one-one function, different elements of (A) must go to different elements of (B). Step 2: The count is (^{6}P_{4}=6\cdot5\cdot4\cdot3=360). Step 3: Repetition of images is not allowed in one-one functions.

Step 1: Each element of (A) has (5) choices in (B). Step 2: For (3) elements, the total number is (5^3=125). Step 3: In total functions, repeated images are allowed.

Step 1: The total number of functions is (2^3=8). Step 2: Non-onto functions send all elements of (A) to only one element of (B), so there are (2) such functions. Step 3: Hence the number of onto functions is (8-2=6).

Step 1: For finite sets of equal size, bijective functions correspond to permutations. Step 2: The number is (4!=24). Step 3: A bijection uses every image exactly once.

Step 1: An invertible function must be both one-one and onto. Step 2: If (a=0), the function becomes constant and is not one-one. Step 3: Therefore the linear function is invertible when (a\neq0).

Step 1: (f(2)=2) and (f(-2)=2), so the function is not one-one. Step 2: Every value (y\geq0) in the codomain is obtained by taking (x=y). Step 3: In modulus functions, opposite signs can give the same output.

Step 1: \(f(2)=\frac{5}{2}\) and \(f\left(\frac{1}{2}\right)=\frac{5}{2}\), so it is not one-one. Step 2: Since \(x+\frac{1}{x}\geq2\) for \(x>0\), negative values are not obtained. Step 3: For such functions, compare (x) and \(\frac{1}{x}\).

Step 1: On (x\geq0), (x^2) is increasing, so different inputs give different outputs. Step 2: Every (y\geq0) is obtained by (x=\sqrt{y}). Step 3: Restricting the domain to ([0,\infty)) makes the square function invertible.

Step 1: From (y=\frac{2x+3}{x-1}), we get (x=\frac{y+3}{y-2}). Step 2: For every (y\neq2), exactly one (x) is obtained. Step 3: Therefore with codomain (\mathbb{R}-{2}), the function is bijective.

Step 1: Write (y=\frac{3x-1}{2x+5}). Step 2: From (2xy+5y=3x-1), we get (x(3-2y)=5y+1). Step 3: Thus (x=\frac{5y+1}{3-2y}), and replacing (y) by (x) gives the inverse.

Step 1: ((g\circ f)(x)=g(f(x))). Step 2: (g(x^2+1)=2(x^2+1)-3=2x^2-1). Step 3: Substitute the inner function first, then apply the outer function.

Step 1: ((f\circ g)(x)=x^2+2). Step 2: ((g\circ f)(x)=(x+2)^2). Step 3: Changing the order of composition generally changes the result.

Step 1: If ((g\circ f)(a_1)=(g\circ f)(a_2)), then (g(f(a_1))=g(f(a_2))). Step 2: Since (g) is one-one, (f(a_1)=f(a_2)), and since (f) is one-one, (a_1=a_2). Step 3: The composition of two one-one functions is one-one.

Step 1: Take any element of (C). Step 2: Since (g) is onto, it is the image of some element of (B), and since (f) is onto, that element of (B) comes from some element of (A). Step 3: Therefore (g\circ f) reaches every element of (C).