Functions

Step 1: Write (x^2+2x+3=(x+1)^2+2). Step 2: Since ((x+1)^2\geq 0), the least value is (2). Step 3: Completing the square is a quick method for finding the range of a quadratic function.

Step 1: Put (y=2x-5). Step 2: Solving for (x) gives (x=\frac{y+5}{2}). Step 3: While finding the inverse, replace (y) by (x) at the end.

Step 1: Each element of (A) has (2) choices in (B). Step 2: For (3) elements, the total number of functions is (2^3=8). Step 3: In counting functions, the base is the number of elements in the codomain.

Step 1: ((g\circ f)(x)) means (g(f(x))). Step 2: Substituting (f(x)=x+3) into (g) gives (g(x+3)=(x+3)^2). Step 3: In composition, changing the order can change the answer.

Step 1: If (x_1<x_2), then (x^3+x) increases. Step 2: A strictly increasing function cannot give the same value for two different inputs. Step 3: Monotonic behaviour is useful for checking one-one functions.

Step 1: A one-one function must give different outputs for different inputs. Step 2: Here (1\neq -1), but (f(1)=1) and (f(-1)=1). Step 3: One counterexample is enough to prove that a function is not one-one.

Step 1: For every (y\geq 0), choosing (x=\sqrt{y}) gives (f(x)=y), so it is onto. Step 2: Since (f(1)=f(-1)), it is not one-one. Step 3: The codomain strongly affects onto behaviour.

Step 1: (x^2+1\geq 1) for every real (x). Step 2: So (0) is in the codomain but is never attained. Step 3: To disprove onto, show one missing element of the codomain.

Step 1: Let (y=\frac{2x+3}{x-1}). Step 2: Putting (y=2) gives (2x+3=2x-2), which is impossible. Step 3: For rational functions, solve in terms of (y) to find excluded values.

Step 1: First find (g(2)=2-4=-2). Step 2: Then (f(-2)=3(-2)+1=-5). Step 3: In composition, apply the inner function first.

Step 1: A one-one function sends distinct elements of (A) to distinct elements of (B). Step 2: (A) has (5) elements but (B) has only (4). Step 3: By the pigeonhole idea, such a one-one function cannot exist.

Step 1: For finite sets of equal size, functions that are one-one and onto are permutations. Step 2: The number of permutations of (4) elements is (4!=24). Step 3: Use (n!) directly in such bijection counting questions.

Step 1: An invertible function must be one-one and onto. Step 2: If (a\neq 0), the linear function takes every real value exactly once. Step 3: If (a=0), it becomes constant and is not invertible.

Step 1: (f(2)=f(-2)), so the function is not one-one. Step 2: No negative real number can be its output, so it is not onto. Step 3: For absolute value functions, check both sign and range.

Step 1: (f^{-1}(8)) means the (x) for which (x^3=8). Step 2: Taking (x=2) gives (2^3=8). Step 3: In inverse values, we find the input from the output.

Step 1: ((f\circ g)(x)=f(g(x))). Step 2: Substitute (g(x)=x^2-1) into (f) to get (2(x^2-1)+1). Step 3: Simplifying gives (2x^2-1).

Step 1: Different values of (n) give different values of (n+1), so it is one-one. Step 2: The number (1) has no preimage in natural numbers. Step 3: For shift functions on natural numbers, the first value is often missed.

Step 1: (x+2) sends different integers to different integers. Step 2: For any (y\in\mathbb{Z}), choose (x=y-2\in\mathbb{Z}). Step 3: Shifts on integers are usually bijective.

Step 1: Since (x^2\geq 0), (1+x^2\geq 1). Step 2: The maximum value is (1) at (x=0), and the function approaches (0) but never reaches it. Step 3: For rational ranges, watch endpoint inclusion carefully.

Step 1: Put (y=\frac{1}{x}). Step 2: This gives (x=\frac{1}{y}). Step 3: Hence the inverse function is the same as the original function.

Step 1: Write (x^2-4x+7=(x-2)^2+3). Step 2: Since ((x-2)^2\geq 0), the minimum value is (3). Step 3: Completing the square quickly gives the minimum of a quadratic function.

Step 1: For an onto function, every element of (B) must be hit by some element of (A). Step 2: (A) has (3) elements but (B) has (4), so this is impossible. Step 3: For finite sets, the domain must not be smaller than the codomain for onto functions.

Step 1: The linear function (kx+7) is one-one when (k\neq 0). Step 2: To fail one-one, its slope must be (0). Step 3: At (k=0), the function becomes constant.

Step 1: (x^5+x) is strictly increasing on the real line, so it is one-one. Step 2: As (x\to\infty), the value goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Step 3: An increasing function covering all real values is bijective.

Step 1: (f(1)=f(-1)), so it is not one-one. Step 2: Negative real numbers are not obtained as (x^4), so it is not onto. Step 3: For even powers, check both symmetry and range.