- f : R → R defined by f(x) = 3 – 4x
- f : R → R defined by f(x) = 1 + x2
Solution:
(i) f : R → R characterized by f(x) = 3 – 4x
On the off chance that x1, x2 ∈ R
f(x1) = 3 – 4×1 and f(x2) = 3 – 4×2
On the off chance that f(x1) = f(x2) x1 = x2 Therefore, f is one-one.
Once more,
f(x) = 3 – 4x or y = 3 – 4x
or then again x = (3-y)/4 in R
f((3-y)/4) = 3 – 4((3-y)/4) = y
f is onto.
Thus f is onto or bijective.
(ii) f : R → R characterized by f(x) = 1 + x2
On the off chance that x1, x2 ∈ R
f(x1) = 1 + ????2 and f(x2) = 1 + ????2
On the off chance that f(x1) = f(x2) ????2 = ????2
1 2
This suggests x1 ≠ x2
Along these lines, f isn’t one-one
Once more, if each component of co-area is picture of some component of Domain under f, with the end goal that f(x) = y
f(x) = 1 + x2
y = f(x) = 1 + x2 or x = ±√1 − ????
Along these lines, f(√1 − ???? ) = 2 – y ≠ y
Along these lines, f isn’t onto or bijective.