Functions

Step 1: The denominator is (x-1). Step 2: The denominator cannot be zero, so (x-1\neq 0), meaning (x\neq 1). Step 3: The domain includes only those real numbers for which the function is defined.

Step 1: For every (x), the function value is always (5). Step 2: So it is a constant function and not one-one, because many inputs have the same image. Step 3: To identify a constant function, check whether the value depends on (x).

Step 1: (f^{-1}(7)) means the value of (x) for which (f(x)=7). Step 2: (x^3-1=7), so (x^3=8) and (x=2). Step 3: For inverse value questions, set the original function equal to the given output.

Step 1: The expression inside the square root must be non-negative. Step 2: (x-3\geq 0), so (x\geq 3). Step 3: For square-root functions, always set the inside expression (\geq 0).

Step 1: Each element of (A) can be mapped to any one of the (n) elements of (B). Step 2: There are (m) independent choices. Step 3: By the multiplication rule, the total number of functions is (n^m).

Step 1: If (a=0), then (f(x)=b) becomes a constant function. Step 2: A constant function cannot be one-one. Step 3: Therefore, a linear function (ax+b) is one-one only when (a\neq 0).

Step 1: ((f\circ g)(x)=f(x+1)=\frac{1}{x+1}). Step 2: The denominator (x+1) must not be zero. Step 3: Hence (x\neq -1), so the domain is (\mathbb{R}-{-1}).

Step 1: (f(1+h)=(1+h)^2-2(1+h)=h^2-1). Step 2: (f(1)=1-2=-1). Step 3: The difference is ((h^2-1)-(-1)=h^2).

Step 1: (x^5+x^3+x) keeps increasing with (x), so it is one-one. Step 2: For very negative (x), values are very negative, and for very positive (x), values are very positive, so all real values occur. Step 3: Increasing odd-degree polynomial functions often have range (\mathbb{R}).

Step 1: Write (y=\frac{x+2}{x-2}). Step 2: (y(x-2)=x+2), so (x(y-1)=2y+2) and (x=\frac{2y+2}{y-1}). Step 3: Replace (y) by (x) at the end to get the inverse.

Step 1: For every (y\geq 0), there is an (x), such as (x=\sqrt{y}), so the function is onto. Step 2: (f(2)=f(-2)=4), so it is not one-one. Step 3: Changing the codomain can change whether a function is onto.

Step 1: On ([0,\infty)), (x^2) is increasing, so distinct inputs give distinct outputs. Step 2: For every (y\geq 0), (x=\sqrt{y}) lies in the domain. Step 3: Restricting the domain can make the same formula one-one.

Step 1: First apply (f(x)=x+1). Step 2: Then (g(f(x))=2(x+1)=2x+2). Step 3: Finally (h(2x+2)=(2x+2)^2=4x^2+8x+4).

Step 1: In an onto function, all (3) elements of (B) must appear as images. Step 2: (A) has only (2) elements, so at most (2) distinct images are possible. Step 3: When the domain is smaller than the codomain, an onto function is impossible.

Step 1: Complete the square: (x^2+2x+2=(x+1)^2+1). Step 2: Since ((x+1)^2\geq 0), the minimum value is (1). Step 3: Thus the range is ([1,\infty)).

Step 1: In a rational function, the denominator cannot be zero. Step 2: Substituting (x=-4) gives (x+4=0). Step 3: Hence (x=-4) is not included in the domain.

Step 1: ((g\circ f)(x)=g(3x-1)). Step 2: (g(3x-1)=\frac{3x-1+1}{3}=x). Step 3: A function that gives (x) for every (x) is the identity function.

Step 1: For every real (x), (x^4\geq 0). Step 2: Therefore (f(x)=x^4+1\geq 1), so a real number like (0) is not attained. Step 3: To test onto, look for values in the codomain that are missed.

Step 1: One-one ensures each image has a unique original element. Step 2: Onto ensures every element of (B) is actually an image. Step 3: Therefore, the inverse function exists from (B) to (A).

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