Question

plz solve this question fast​

Answer 1

Step-by-step explanation:

case 1: f(x) = 3x - 2

put -ve values of (x)

f(-1) = -3 -2

= -5

f(-2) = -6 -2

= -8

case 2: x = 0

f(x) = 1

case 3: f(x) = 4x + 1

put +ve values of (x)

f(1) = 4+1

= 5

f(2) = 4×2 +1

= 9

so, these are some functions ......

I think this will help u........

Answer 2

Given :-

• F: R→ R

• f(x) = 3x-2 , x < 0

• f(x) = 1 , x = 0

• f(x) = 4x +1 , x > 0

Solution :-

1st case → f( 1)

Here x = 1 , x is greater than 0 .

→ f(x) = 4x +1

→ f(1) = 4(1) +1

→ f(1) = 4+1

→ f(1) = 5

2nd case → f(-1)

Here x = -1 , x is smaller than 0 .

→ f(x) = 3x -2

→ f(-1) = 3(-1) - 2

→ f(-1) = -3 -2

→ f(-1) = -5 .

3rd case → f(0)

Here x = 0 , x is equivalent to 0

→ f(0) = 1 .

4th case → f(2)

Here x = 2 , x is greater than 0 .

→ f(x) = 4x +1

→ f(2) = 4(2) +1

→ f(2 ) = 8 +1

→ f(2) = 9