Hi.. :)
Q5) f(x) can be written as:
= x^6 for x>1 and x < -1
= x^3 for -1 ≤ x ≤ 1.
For x < -1 and x > 1, f(x) is a polynomial. Hence, continuous.
For x = 1, LHL = RHL = 1. Hence, continuous.
For x = -1, LHL = 1 and RHL = -1. Therefore, not continuous at x = -1.
Q6) Does not exist.
The range of h(x) is [0,∞). To be precise, the range is positive integers only.
Therefore, there is not value of x which will be mapped to -1. Hence, h inverse at -1 does not exist.
:)