Functions

Step 1: First (f) sends an element of (A) to (B). Step 2: Then (g) sends that element of (B) to (C). Step 3: Hence the composite function (g\circ f) maps from (A) to (C).

Step 1: ((g\circ f)(1)=g(f(1))). Step 2: (f(1)=2\cdot1+1=3). Step 3: (g(3)=3^2=9), so the value is (9).

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

Step 1: (f(g(x))=f(x-3)=(x-3)+3=x). Step 2: (g(f(x))=g(x+3)=(x+3)-3=x). Step 3: Both composites give the identity function, so they are inverses.

Step 1: In (x+1), (x) can take all real values. Step 2: For any real (y), choosing (x=y-1) gives (f(x)=y). Step 3: Therefore the range is all of (R).

Step 1: For every real (x), (x^2\ge 0). Step 2: Therefore (x^2+1\ge 1). Step 3: The minimum value is (1), so the range is ([1,\infty)).

Step 1: In a many-one function, different inputs may have the same image. Step 2: For (f(x)=x^2), (f(1)=f(-1)=1). Step 3: Hence (x^2) is many-one, while the other given simple functions are one-one.

Step 1: The images of (1) and (2) are different, (3) and (4), so the function is one-one. Step 2: The element (5) of (B) is not an image of any element. Step 3: Hence it is not onto.

Step 1: Both elements (a) and (b) of (B) appear as images, so the function is onto. Step 2: Both (1) and (2) map to (a), so it is not one-one. Step 3: Check onto and one-one separately.

Step 1: Put (x=3) in the given function. Step 2: (f(3)=4\cdot3-5=12-5=7). Step 3: In a linear function, multiply first and then add or subtract.

Step 1: Put (x=5). Step 2: (f(5)=5^2-2\cdot5=25-10=15). Step 3: Calculate the square and multiplication carefully.

Step 1: On putting (x=7), the numerator becomes (7+1=8). Step 2: Hence (f(7)=\frac{8}{2}=4). Step 3: In fraction-based functions, simplify the numerator first.

Step 1: Put (x=-1). Step 2: (f(-1)=3(-1)^2+1=3\cdot1+1=4). Step 3: Remember that the square of a negative number is positive.

Step 1: (A) is the domain of the function. Step 2: Elements of the domain are treated as inputs. Step 3: The image of an input lies in (B).

Step 1: In an ordered pair, the first component is the input and the second component is the image. Step 2: Since ((4,7)) is given, the image of (4) is (7). Step 3: Do not interchange the first and second components while reading a pair.

Step 1: For being a function, every input must have exactly one image. Step 2: Here (1,2,3) each have one image. Step 3: Two different inputs may have the same image; that does not stop it from being a function.

Step 1: In a function, one input cannot have two different images. Step 2: In the first option, (1) has images (2) and (3). Step 3: Therefore the first option is not a function.

Step 1: Apply the function to every element of (A). Step 2: (f(1)=3), (f(2)=4), and (f(3)=5). Step 3: Therefore (f(A)={3,4,5}).

Step 1: For (f(a+b)), replace (x) by the whole expression (a+b). Step 2: This gives (f(a+b)=5(a+b)). Step 3: Put the entire input in brackets.

Step 1: For (f(a)), replace (x) by (a). Step 2: (x^2-1) becomes (a^2-1). Step 3: While changing the symbol, keep the power correctly.