Functions

Step 1: ((f\circ g)(x)=f(g(x))). Step 2: Put (g(x)=x+5) into (f). Step 3: (f(x+5)=3(x+5)-2=3x+13).

Step 1: ((f\circ g)(x)=f(x-1)=(x-1)^2+2). Step 2: ((g\circ f)(x)=g(x^2+2)=x^2+1). Step 3: The expressions are generally different, so order is important in composition.

Step 1: In (7x+2), different (x) values give different outputs, so the function is one-one. Step 2: For any (y\in R), choose (x=\frac{y-2}{7}). Step 3: Thus every real (y) is an image, so the function is onto.

Step 1: (f(2)=f(-2)=-1), so the function is not one-one. Step 2: Since (x^2-5\ge-5), a number like (-6) cannot be an image. Step 3: Hence it is not onto either.

Step 1: For different (x), (x-1) is different, so the function is one-one. Step 2: For every (y\ge0), (x=y+1), which lies in the domain ([1,\infty)). Step 3: Hence the function is onto as well.

Step 1: For every (y\ge4), (x=\sqrt{y-4}) or (x=-\sqrt{y-4}) can be chosen, so it is onto. Step 2: (f(1)=f(-1)=5), so it is not one-one. Step 3: Changing the codomain can change onto status.

Step 1: The expression inside the square root, (2x-6), must be non-negative. Step 2: From (2x-6\ge0), we get (x\ge3). Step 3: At (x=3), the value is (0), so (3) is included.

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

Step 1: The denominator is (x^2-16=(x-4)(x+4)). Step 2: The denominator cannot be zero, so (x\ne4) and (x\ne-4). Step 3: Even if (x-4) cancels, (x=4) is not valid in the original function.

Step 1: To find the preimage, solve (|x+1|=4). Step 2: This gives (x+1=4) or (x+1=-4). Step 3: Thus (x=3) or (x=-5), so the preimage is ({-5,3}).

Step 1: The images of (1,2,3,4) are distinct, so the function is one-one. Step 2: (t) is in the codomain but is not an image. Step 3: Therefore the function is not onto.

Step 1: Total functions are (2^4=16). Step 2: There are two non-onto functions, where all elements go only to (a) or only to (b). Step 3: Therefore onto functions are (16-2=14).

Step 1: In a one-one function, the two inputs must have distinct images. Step 2: The first input has (5) choices and the second has (4) choices. Step 3: Total one-one functions are (5\cdot4=20).

Step 1: In an onto function, every element of the codomain must be an image. Step 2: Here the domain has (3) elements, but the codomain has (4). Step 3: Three inputs cannot cover four different codomain elements, so no onto function exists.

Step 1: To find (f^{-1}(10)), solve (x^3+2=10). 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{3x-2}{4}). Step 2: Then (4y=3x-2), so (x=\frac{4y+2}{3}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\frac{4x+2}{3}).

Step 1: For an inverse function to exist, the original function must be one-one. Step 2: (f(0)=0) and (f(-2)=0), while (0\ne-2). Step 3: Since one image has two preimages, the inverse is not a well-defined function.

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

Step 1: Write (x^2+6x+10=(x+3)^2+1). Step 2: Since ((x+3)^2\ge0), (f(x)\ge1). Step 3: The minimum value is (1), attained at (x=-3).

Step 1: Write the function as ((x+3)^2+1). Step 2: A square is always (0) or more. Step 3: Therefore the least value is (1), and the range is ([1,\infty)).

Step 1: Assume (\frac{2x}{x-1}=2). Step 2: This gives (2x=2x-2), which is impossible. Step 3: Therefore (2) is not in the range of this function.

Step 1: Write (y=\frac{2x}{x-1}). Step 2: From (y(x-1)=2x), we get (x=\frac{y}{y-2}), if (y\ne2). Step 3: Hence every real (y) is possible except (2).

Step 1: Bijective means both one-one and onto. Step 2: The composite of two one-one functions is one-one, and the composite of two onto functions is onto. Step 3: Therefore (g\circ f) is bijective.

Step 1: Suppose (f(a_1)=f(a_2)). Step 2: Then (g(f(a_1))=g(f(a_2))), so ((g\circ f)(a_1)=(g\circ f)(a_2)). Step 3: Since (g\circ f) is one-one, (a_1=a_2), so (f) is one-one.

Step 1: Inverse functions undo each other's action. Step 2: Applying (f) first and then (g) brings the original value back. Step 3: Therefore ((g\circ f)(x)=x).