If a function f: R → R be defined by f(x) = {(3x-2,x < 0)(1,x = 0)(4

1.	If a function f: R → R be defined by f(x) = {(3x-2,x < 0)(1,x = 0)(4x +1,x > 0)\(f(x) =\begin{cases}3x-2&, \quad x<0\\1&, \quad x=0\\4x+1&,\quad x>0\end{cases}\)
Answer» Given, f(x) = {(3x-2,x < 0)(1,x = 0)(4x +1,x > 0) We need to find, f(1), f(–1), f(0) and f(2). When x > 0, f(x) = 4x + 1 Substituting x = 1 in the above equation, we get f(1) = 4(1) + 1 ⇒ f(1) = 4 + 1 ∴ f(1) = 5 When x < 0, f(x) = 3x – 2 Substituting x = –1 in the above equation, we get f(–1) = 3(–1) – 2 ⇒ f(–1) = –3 – 2 ∴ f(–1) = –5 When x = 0, f(x) = 1 ∴ f(0) = 1 When x > 0, f(x) = 4x + 1 Substituting x = 2 in the above equation, we get f(2) = 4(2) + 1 ⇒ f(2) = 8 + 1 ∴ f(2) = 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) = {(3x-2,x < 0)(1,x = 0)(4x +1,x > 0)\(f(x) =\begin{cases}3x-2&, \quad x<0\\1&, \quad x=0\\4x+1&,\quad x>0\end{cases}\)

Answer»

Given,

f(x) = {(3x-2,x < 0)(1,x = 0)(4x +1,x > 0)

We need to find,

f(1), f(–1), f(0) and f(2).

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

Substituting x = 1 in the above equation, we get

f(1) = 4(1) + 1

⇒ f(1) = 4 + 1

∴ f(1) = 5

When x < 0, f(x) = 3x – 2

Substituting x = –1 in the above equation, we get

f(–1) = 3(–1) – 2

⇒ f(–1) = –3 – 2

∴ f(–1) = –5

When x = 0, f(x) = 1

∴ f(0) = 1

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

Substituting x = 2 in the above equation, we get

f(2) = 4(2) + 1

⇒ f(2) = 8 + 1

∴ f(2) = 9

Thus,

f(1) = 5,

f(–1) = –5,

f(0) = 1 and f(2) = 9.