Functions

Step 1: (f(g(x))=f(2x)=2x+1). Step 2: (g(f(x))=g(x+1)=2x+2). Step 3: Composition of functions is not generally commutative.

Step 1: For (x\geq0), the value is (\frac{x}{1+x}), which approaches (1) but never reaches it. Step 2: For (x<0), the value lies between (-1) and (0). Step 3: Combining both parts gives the range ((-1,1)).

Step 1: Put (y=2x^3-1). Step 2: Then (y+1=2x^3), so (x=\sqrt[3]{\frac{y+1}{2}}). Step 3: In inverse questions, isolate (x) first and then change the variable.

Step 1: For the inverse to be a function, the original function must be one-one. Step 2: (f(1)=f(-1)=1), so one output has two inputs. Step 3: Thus the inverse relation assigns two values and is not a function.

Step 1: (x^2\geq0) for every real (x). Step 2: Hence (x^2+4\geq4), and (4) is attained at (x=0). Step 3: While writing range, check whether the endpoint is included.

Step 1: The range of (\cos x) is ([-1,1]). Step 2: The codomain (\mathbb{R}) contains (2), but (\cos x=2) is impossible. Step 3: To test onto, look for values in the codomain that are missed.

Step 1: To show a function is not one-one, find two different inputs with the same output. Step 2: (f(0)=0) and (f(\sqrt{3})=(\sqrt{3})^3-3\sqrt{3}=0). Step 3: A suitable counterexample is enough to disprove one-one behaviour.

Step 1: (e^x) is strictly increasing on the real line, so it is one-one. Step 2: Its values are positive, and every positive number can be written as (e^x). Step 3: With codomain ((0,\infty)), it becomes bijective.

Step 1: (e^x) is increasing, so it gives different outputs for different inputs. Step 2: It never becomes zero or negative, so it does not cover all of (\mathbb{R}). Step 3: The same formula can have different properties with a different codomain.

Step 1: (\tan x) is not defined where (\cos x=0). Step 2: For example, at (x=\frac{\pi}{2}), (\tan x) is not defined. Step 3: A function must assign a value to every element of its domain.

Step 1: Let (y=\frac{3x+1}{x+2}). Step 2: Putting (y=3) gives (3x+1=3x+6), which is impossible. Step 3: In linear fractional functions, the horizontal limiting value is often the missed value.

Step 1: The images of the three elements of (A) are all distinct. Step 2: The outputs include (1,2,3), so the function is onto. Step 3: For finite equal sets, distinct images imply bijection.

Step 1: An absolute value is never negative. Step 2: At (x=2), (|x-2|=0). Step 3: The minimum of an absolute value function occurs when the inside expression becomes zero.

Step 1: (\lfloor 1.2\rfloor=\lfloor 1.8\rfloor=1), so it is not one-one. Step 2: Its values are always integers, so real values like (0.5) are missed. Step 3: For step functions, test one-one and onto separately.

Step 1: For any real (a), this is a quadratic function. Step 2: On all of (\mathbb{R}), a quadratic takes equal values on opposite sides of its vertex. Step 3: Therefore it cannot be one-one on the entire real domain.

Step 1: Suppose ((g\circ f)(x_1)=(g\circ f)(x_2)). Step 2: Since (g) is one-one, (f(x_1)=f(x_2)); since (f) is one-one, (x_1=x_2). Step 3: The composition of one-one functions is one-one.

Step 1: Take any (c\in C). Step 2: Since (g) is onto, some (b\in B) satisfies (g(b)=c); since (f) is onto, some (a\in A) satisfies (f(a)=b). Step 3: Hence ((g\circ f)(a)=c), so the composition is onto.

Step 1: The denominator is (x-1), so (x=1) is removed from the domain. Step 2: Putting (y=1) in (y=\frac{x+1}{x-1}) gives (x+1=x-1), impossible. Step 3: For rational functions, check domain and missed range values separately.

Step 1: (f(x)=2x-5) is linear with non-zero slope, so distinct inputs give distinct outputs. Step 2: For any (y\in\mathbb{R}), (x=\frac{y+5}{2}) is real, so every real output is attained. Step 3: In exams, remember that (ax+b) with (a\neq0) is both one-one and onto from (\mathbb{R}) to (\mathbb{R}).

Step 1: Since (x^2\ge 0), we get (x^2+1\ge 1). Step 2: The codomain (\mathbb{R}) contains (0), but (f(x)=0) is impossible for real (x). Step 3: To test onto, look for a codomain value that is never produced by the function.