f(2) - f(4)
f(2) = 2x - 1
= 2(2) - 1 = 3
f(4) = 3x² - 10
= 3(42) - 10 = 38
\(\therefore\) f(2) - f(4) = 3 - 38 = 35

\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \)
For f(1) & f(3)
Since 1 & 3 lies between -5 ≤ x ≤2, We take, f(x) = x +5
f(1) = 1 + 5 = 6
f(-3) = -3 +5 = 2
For f(4): Since 4 lies between 2 < x < 6 We take, f(x) = x -1
f(4) = 4 -1 = 3
For f(-6): Since -6 lies between -7≤ x < -5 ,We take, f(x) = x² + 2x +1
f(-6) = (-6)² + 2(-6) + 1 = 36 -12 + 1 = 24 +1 = 25
Now ,\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \) =\(\frac{ 4(2) +2(3) }{ 25 - 3(6)}\)
= \(\frac{ 8 + 6 }{ 25 -18}\)
= \(\frac{14 }{ 7}\) = 2
Hence, the value of \(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \) = 2

f(-7) = x² + 2x + 1
= (-7)² + 2(-7) + 1
= 49 - 14 + 1 = 36
f(3) = x + 5 = -3 + 5 = 2
f(-7) - f(-3) = 36 + 2 = 38

A = {1, 2, 3, 4} ; B = {2, 5, 8,11,14}; f(x) = 3x − 1
f(1) = 3(1) –1 = 3 – 1 = 2; f(2) = 3(2) –1 = 6 –1 = 5
f(3) = 3(3) –1 = 9 –1 = 8; f(4) = 4(3) –1 = 12 –1 = 11
(i) Arrow diagram
Let us represent the function f :A ⟶ B by an arrow diagram  

(ii) Table form
The given function f can be represented in a tabular form as given below

(iii) Set of ordered pairs
The function f can be represented as a set of ordered pairs as
f = {(1,2),(2,5),(3,8),(4,11)}
(iv) Graphical form
In the adjacent xy -plane the points
(1,2), (2,5), (3,8), (4,11) are plotted (Fig.1.20).

Given: A = {1,2,3} , B = {2,3,5} , C = {3,4} ,D = {1,3,5}
\(A\cap C\) = {3}
\(B\cap D\) = {3,5}
\((A\cap C)\times(B\cap D)=\{ 3\} \times \{ 3,5\} \)
= {(3,3),(3,5)}          ..(1)
A x B = {(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5)}
C x D = {(3,1),(3,3),(3,5),(4,1),(4,3),(4,5)}
\((A\times B)\cap (C\times D)\)  = {(3,3),(3,5)}                ....(2)
From (1) and (2),it is clear that
(A ∩ C) x (B ∩ D) = (A x B) ∩ (C x D)
Hence it is true.

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

An arrow diagram
 

A table

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)

A = {1, 2, 3}, B = {4, 5, 6, 7}; f = {(1, 4),(2, 5),(3, 6)}
Then f is a function from A to B and for different elements in A, there are different images in B. Hence f is one–one function. Note that the element 7 in the co-domain does not have any pre-image in the domain. Hence f is not onto.
Therefore f is one–one but not an onto function.

Pictorial representation of R . From the diagram, we see that for each x \(\in \) X, there exists only one y \(\in \) Y. Thus all elements in X have only one image in Y. Therefore R is a function Domain X = {1, 2, 3, 4}; Co-domain Y = {2, 3, 6, 8,10}; Range of f = {2, 4, 6, 8}.