Question -
Answer -
Given:
Let us find f (1), f (тАУ1), 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
= 4 + 1
= 5
When x < 0, f(x) = 3x тАУ 2
Substituting x = тАУ1 in the above equation, we get
f (тАУ1) = 3(тАУ1) тАУ 2
= тАУ3 тАУ 2
= тАУ5
When x = 0, f(x) = 1
Substituting x = 0 in the above equation, we get
f (0) = 1
When x > 0, f(x) = 4x + 1
Substituting x = 2 in the above equation, we get
f (2) = 4(2) + 1
= 8 + 1
= 9
тИ┤ f (1) = 5, f (тАУ1) = тАУ5, f (0) = 1 and f (2) = 9.