A semester course in trigonometry by Marcel Finan

By Marcel Finan

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Then for all x in the common domain of these two functions we define new functions as follows: • • • • Sum: (f + g)(x) = f (x) + g(x). Difference: (f − g)(x) = f (x) − g(x). Product: (f · g)(x) = f (x) · g(x). (x) provided that g(x) = 0. 1 √ Let f (x) = x + 1 and g(x) = x + 3. Find the common domain and then find a formula for each of the functions f + g, f − g, f · g, fg . Solution. The domain of f (x) consists of all real numbers whereas the domain of g(x) consists of all numbers x ≥ 3. Thus, the common domain is the interval [−3, ∞).

A) f (x) = |x|, if x ≤ 1 2, if x > 1. (b) g(x) =   4, if x ≤ −1 x2 , if −1 < x < 1  −x + 5, if x ≥ 1. 8 Use the vertical line test to determine which of the following graphs are graphs of functions. 9 Use the indicated graphs to indentify the intervals over which the function 37 is increasing, constant, or decreasing. 10 Use the horizontal line test to determine which of the following functions are one-to-one. 11 A bus was purchased for $80,000. Assumeing the bus depreciates linearly at a rate of $6,500 per year for the first 10 years, write the value v of the bus as a function of the time t (measured in years) for 0 ≤ t ≤ 10.

We call such a function an even function. 3 (a) Using a graphing calculator show that the function f (x) = (x − x3 )2 is even. (b) Now show that f (x) is even algebraically. Solution. (a) The graph of f (x) is symmetric about the y-axis so that f (x) is even. 3. 3 (b) Since f (−x) = (−x − (−x)3 )2 = (−x + x3 )2 = [−(x − x3 )]2 = (x − x3 )2 = f (x) then f (x) is even. e, f (−x) = −f (x), then the graph of f (x) is symmetric about the origin. In this case, we say that f (x) is odd. Alternatively, since f (x) = −f (−x), if the graph of a function is reflected first across the y-axis and then across the x-axis and you get the graph of f (x) again then the function is odd.

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