If a function f: R → R be defined by f(x) = \(\begin{cases}3x - 2

1.	If a function f: R → R be defined by f(x) = \(\begin{cases}3x - 2, & \quad \text{ } x < 0 \text{ }\\1, & \quad \text{ } x = 0\text{ }\\ 4x + 1, & \quad \text{ } x > 0\text{ }\end{cases}\)Find: f (1), f (–1), f (0), f (2).
Answer» Given as Let us find the f(1), f(–1), f(0) and f(2). If x > 0, f (x) = 4x + 1 By substituting x = 1 in the above equation, we get f (1) = 4(1) + 1 = 4 + 1 = 5 If x < 0, f(x) = 3x – 2 By substituting x = –1 in the above equation, we get f (–1) = 3(–1) – 2 = –3 – 2 = –5 If x = 0, f(x) = 1 By, substituting x = 0 in the above equation, we get f(0) = 1 If x > 0, f(x) = 4x + 1 By substituting x = 2 in the above equation, we get f(2) = 4(2) + 1 = 8 + 1 = 9 Thus f(1) = 5, f(–1) = –5, f(0) = 1 and f(2) = 9.

If a function f: R → R be defined by f(x) = \(\begin{cases}3x - 2, & \quad \text{ } x < 0 \text{ }\\1, & \quad \text{ } x = 0\text{ }\\ 4x + 1, & \quad \text{ } x > 0\text{ }\end{cases}\)Find: f (1), f (–1), f (0), f (2).

Answer»

Given as

Let us find the f(1), f(–1), f(0) and f(2).

If x > 0, f (x) = 4x + 1

By substituting x = 1 in the above equation, we get

f (1) = 4(1) + 1

= 4 + 1

= 5

If x < 0, f(x) = 3x – 2

By substituting x = –1 in the above equation, we get

f (–1) = 3(–1) – 2

= –3 – 2

= –5

If x = 0, f(x) = 1

By, substituting x = 0 in the above equation, we get

f(0) = 1

If x > 0, f(x) = 4x + 1

By substituting x = 2 in the above equation, we get

f(2) = 4(2) + 1

= 8 + 1

= 9

Thus f(1) = 5, f(–1) = –5, f(0) = 1 and f(2) = 9.