(a) f(x) = ln(1 + x squared)
(b) f(x) = 1 / (over) x-1 + 1 / (over) x-2
See attachment. It includes pictures. Text is pasted here for reference.
If you have a function f(x) and you want to know its domain and range, there are two questions you need to ask:
1. For which values of x does f(x) exist? (domain)
2. What values of f(x) are possible? (range)
We usually answer question 1 by thinking of what values of x wouldn't allow us to find f(x). Finding the range is a little more complicated.
(a) f(x) = ln(1 + x^2)
1. Domain: What problems can we have evaluating the ln function? We know that it doesn't exist at 0 or at negative numbers. But will (1+x^2) ever be 0 or negative? Well look at x2. It's never any smaller than 0:
so when we add 1, we'll never get anything smaller than 1:
So we'll always be taking the ln of something 1 or bigger. That means ln(1+x^2) exists for all real x, since we don't need to worry about those negative numbers. The domain is all real numbers.
2. Range: let's see, what values can ln take on? All real values, ...
This solution shows how to find the domain and range of two nonlinear functions, with explanatory pictures.