A Graph f = {(x, f(x) / x \(\epsilon \) A}
{(0, 1), (1, 3), (2, 5), (3, 7)}

An arrow diagram
 

A table

It is not a function as the vertical line PQ cuts the graph at two points

A = {1,2,3},B =  {1,3,5, 7,9}
f(x) = 2x + 1
f(0)) = 2(0) + 1 = 1
f(1) =  2(1) + 1=3
f(2) = 2(2) + 1 = 5
f(3) = 2(3) + 1 = 7
(i) A set of ordered pairs.
f = {(0, 1), (1, 3), (2, 5), (3, 7)}
(ii) A table

(ii) An arrow diagram

(iv) A Graph f = \(\{(x, f(x) / x \in A\}\)
= {(0, 1), (1, 3), (2, 5), (3, 7)}

Domain of R = {1,2,3,4}
Co-domain of R = B = {-1, 2, 3,4,5,6, 7, 9, 10, 11,12}
Range of R = {3, 6,10, 9}

f(x) = 2x + 3, g(x) = 1 - 2x, h(x) = 3x
Now, (f o g)(x) = f(g(x)) = f(1 - 2x) = 2(1 - 2x) + 3 = 5 - 4x
Then, (f o g) o h(x) = (f o g)(3x) = 5 - 4(3x) = 5 - 12x..(1)
(g o h)(x) = g(h(x)) = g(3x) = 1 - 2(3x) = 1 - 6x
So, f o (g o h)(x) = f(1 - 6x) = 2(1 - 6x) + 3 = 5 - 12x...(2)
From (1) and (2), we get (f o g) oh = f o (g o h)

f o f(k) = f(f(k))
= 2(2k - 1) -1 = 4k - 3
Thus, f o f(k) = 4k - 3
But, it is given that f o f(k) = 5
Therefore 4k - 3 = 5 ⇒ k = 2

f(x) = 3 x -2, g(x) = 2x + k
f o g = f(g(x)) = f(2x + k) = 3(2x + k) - 2 = 6x + k - 2
Thus, f o g(x) = 6x + 3k - 2
g o f(x) = g(3x - 2) = 2(3x - 2) + k
Thus, g o f(x) = 6x - 4 + k
Given that f o g = g o f
Therefore, 6x + 3k-2 = 6z - 4 + k
6x - 6x + 3k - k = -4 + 2 ⇒ -1

We set f₂(x) = 2x² - 5x + 3 and f₁(x) =\(\sqrt { x } \)
Then, f(x) = \(\\ \sqrt { 2x^{ 2 }-5x+3 } =\sqrt { { f }_{ 2 }(x) } \)
= f₁{f₂(x)} = f₁f₂(x)

f(x) = 2x + 1, g(x) = x² - 2
f o g(x) = f(g(x)) = f(x² - 2) = 2(x² - 2) + 1 = 2x² - 3
g o f(x) = g(f(x)) = g(2x + 1) = (2x + 1)² - 2 = 4x² + 4x - 1
Thus f o g = 2x² - 3, g o f = 4x² + 4x - 1. From the above, we see that f o g ≠ g o f.

R₁ = {(–5, a), (1, a), (3, b)}
We may represent the relation R₁ in an arrow diagram 
R₁ is not a function as 4 \(\in\) X does not have an image in y.

f(x) = x² - 2 where x \(\in \){ -2, -1, 0, 3}
(i) f( -2) = ( -2)² - 2 = 2; f( -1) = ( -1)² - 2 = -1
f(0) = (0)² - 2 = - 2 ; f(3) = (3)² - 2 = 7
Therefore, f = {(-2, 2), (-1, -1), (0, -2), (3, 7)}
(ii) We note that each element in the domain of f has a unique image. Therefore f is a function.

gff(x) = g[f{f(x)}] (This means "g of f of f of x")
= g[f(3x + 1)] = g[3(x + 1) + 1] = g(9x + 4)
g(9x + 4) = [(9x + 4) + 3] = 9x + 7
fgg(x) = f[g{g(x)}] (This means " f of g of g of x")
= f[g(x + 3)] = f[(x + 3) + 3] = f(x + 6)
f(x + 6) = [3(x + 6) + 1] = 3x + 19
These two quantities being equal, we get 9x + 7 = 3x + 19. Solving this equation we obtain x = 2.

(i) x = 1, we get
f(1) = 2(1) - (1)² = 2 - 1 = 1
(ii) x = x + 1, we get
f(x + 1) = 2(x + 1) - (x + 1)² = 2x + 2 - (x² + 2x + 1) = -x² + 1
(iii) f(x) + f(1) = (2x - x²) + 1= - x² + 2x + 1
[Note that f(x) + f(1) ≠ f(x + 1). In general f(a + b) is not equal to f(a) + f(b)]