Functions

Step 1: (f^{-1}(8)) means the value of (x) for which (f(x)=8). Step 2: From (3x-4=8), we get (3x=12), so (x=4). Step 3: To find an inverse value, first equate the original function to the given value.

Step 1: Write (f(x+1)=(x+1)^2+2(x+1)). Step 2: Simplifying gives (x^2+4x+3), and (f(x)=x^2+2x). Step 3: Subtracting gives (2x+3).

Step 1: The denominator of the fraction is (x+2). Step 2: For the function to be defined, (x+2\ne0), so (x\ne-2). Step 3: The value that makes the denominator zero is removed from the domain.

Step 1: Write (x^2-2x+3=(x-1)^2+2). Step 2: Since ((x-1)^2\ge0), we have (f(x)\ge2). Step 3: The minimum value is (2), so the range is ([2,\infty)).

Step 1: ((g\circ f)(2)=g(f(2))). Step 2: (f(2)=2\cdot2+1=5). Step 3: (g(5)=5^2-1=24), so the answer is (24).

Step 1: ((f\circ g)(x)=f(g(x))). Step 2: Put (g(x)=2x-1) into (f). Step 3: (f(2x-1)=2x-1+3=2x+2).

Step 1: ((f\circ g)(x)=f(x+1)=(x+1)^2). Step 2: ((g\circ f)(x)=g(x^2)=x^2+1). Step 3: These expressions are generally not equal, so order matters.

Step 1: (5x-7) gives different values for different (x), so it is one-one. Step 2: For any real (y), choosing (x=\frac{y+7}{5}) gives (f(x)=y). Step 3: Therefore it is onto as well.

Step 1: (f(1)=f(-1)=2), so it is not one-one. Step 2: Since (x^2+1\ge1), a real number like (0) cannot be an image. Step 3: Therefore it is not onto either.

Step 1: On ([0,\infty)), (x^2) is increasing, so different (x) values give different images. Step 2: For every (y\ge0), (x=\sqrt{y}) lies in the domain. Step 3: Hence the function is both one-one and onto.

Step 1: For every (y\ge0), (x=\sqrt{y}) or (x=-\sqrt{y}) gives an input, so the range equals the codomain. Step 2: (f(1)=f(-1)=1), so it is not one-one. Step 3: Changing the codomain can change onto status.

Step 1: The expression inside the square root, (x-2), must be non-negative. Step 2: (x-2\ge0) gives (x\ge2). Step 3: At (x=2), the value is (0), so (2) is included.

Step 1: For the square root, (x-1\ge0) is needed. Step 2: But (\sqrt{x-1}) is in the denominator, so it cannot be zero. Step 3: Thus (x-1>0), so the domain is ((1,\infty)).

Step 1: The denominator is (x^2-9=(x-3)(x+3)). Step 2: The denominator must not be zero, so (x\ne3) and (x\ne-3). Step 3: Even if (x+3) cancels algebraically, (x=-3) is not allowed in the original function.

Step 1: For (f^{-1}({3})), solve (|x-2|=3). Step 2: This gives (x-2=3) or (x-2=-3). Step 3: Hence (x=5) or (x=-1), so the preimage is ({-1,5}).

Step 1: The images of (1,2,3) are distinct, so the function is one-one. Step 2: (d) is in the codomain but is not an image of any element. Step 3: Therefore the function is not onto.

Step 1: Total functions are (2^3=8). Step 2: Non-onto functions are those sending all elements only to (a) or only to (b), so there are (2). Step 3: Hence onto functions are (8-2=6).

Step 1: In a one-one function, the three inputs must have distinct images. Step 2: The first input has (5) choices, the second has (4), and the third has (3). Step 3: Total functions are (5\cdot4\cdot3=60).

Step 1: In an onto function, every element of the codomain must be an image. Step 2: Here the domain has only (2) elements, while the codomain has (3). Step 3: Two inputs cannot cover three distinct codomain elements, so no onto function exists.

Step 1: To find (f^{-1}(7)), solve (x^3-1=7). Step 2: (x^3=8), so (x=2). Step 3: For a cubic function, the real cube root can be taken directly.

Step 1: Write (y=\frac{2x+1}{3}). Step 2: Then (3y=2x+1), so (x=\frac{3y-1}{2}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\frac{3x-1}{2}).

Step 1: For an inverse function to exist, the original function must be one-one. Step 2: Here (f(1)=f(-1)=1), while (1\ne-1). Step 3: Since one image has two preimages, the inverse is not a well-defined function.

Step 1: To find the preimage, solve (|x|=2). Step 2: This gives (x=2) or (x=-2). Step 3: Therefore (f^{-1}({2})={-2,2}).

Step 1: Write (x^2-4x+5=(x-2)^2+1). Step 2: Since ((x-2)^2\ge0), (f(x)\ge1). Step 3: The minimum value is (1), attained at (x=2).

Step 1: Write the function as ((x-2)^2+1). Step 2: A square is zero or positive. Step 3: Therefore the least value is (1), and the range is ([1,\infty)).