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Let f(x) = 1/(x-3) and g(x) = sin2 (x).

(i) Find the domain and range of f.
(ii) Find the domain and range of g.
(iii) Find a formula for (f ◦ g)(x). Hence find (f ◦ g)(1).

(i) all real numbers except where the denominator is zero

(ii) sin and cos have a domain of all real numbers
(f◦g)(x) = f(g(x)) = 1/(sin^2(x) - 3)
so plug in x=1

Did you know?

Did you know that the domain of a function refers to all possible input values for which the function is defined, while the range refers to the set of all possible output values? In the case of function f(x) = 1/(x-3), the domain is all real numbers except for x = 3, since it would result in division by zero. The range of f(x) is all real numbers except 0, as f(x) can take on any positive or negative value. Similarly, for function g(x) = sin^2(x), the domain is all real numbers, since sine function is defined for all x values. The range of g(x) is [0, 1], as sin^2(x) can never be negative and is bounded by 0 and 1. Now, (f ◦ g)(x) represents the composition of f and g, meaning that we substitute the result of g(x) into f(x). By substituting g(x) = sin^2(x) into f(x) = 1/(x-3), we get (f ◦ g)(x) = 1/(sin^2(x) - 3). To find (f ◦ g)(1), we substitute x = 1 into the formula, giving us (f ◦ g)(1) = 1/(sin^2(1) - 3).