Functions+Notes+_updated_.pdf

Solution the domain of f is r because f is defined

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Solution: The domain of f is R - { 0 } because f is defined for every real number except 0. If x ≥ - 1, then | 1 + x | = 1 + x , and so the formula for f ( x ) simplifies to f ( x ) = | 1 + x | - 1 x = 1 + x - 1 x = x x = 1 . If x < - 1, then | 1 + x | = - (1 + x ) = - 1 - x , and we have that f ( x ) = | 1 + x | - 1 x = - 1 - x - 1 x = - 2 - x x = - 1 - 2 /x. Therefore: f ( x ) = - 1 - 2 /x if x < - 1 1 if x ≥ - 1 and x 6 = 0 To sketch the graph of f , notice for x < - 1 the graph of y = - 1 - 2 /x is the graph of y = - 1 /x compressed vertically and shifted down 1 unit. -1 0 1 -1 x y y = - 1
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38 J. S´ anchez-Ortega Example 2.29. Find a formula for function f ( x ) graphed below. What is the domain of f ? Sketch the graph of the functions y = f ( x - 1) and y = f ( x - 1) + 2 . -1 -2 -3 2 3 0 -1 1 2 x y Solution: First of all, notice that the domain of f is [ - 3 , 3) since f is only defined on that interval. The graph consists of parts of three lines. For the part, - 3 < x < - 1, the line has slope m = - 1 - 0 - 1 - ( - 2) = - 1 , and y -intercept b = 0 - ( - 1)( - 2) = - 2 . So its equation is y = - x - 2 = - ( x + 2). The middle section is the line y = x for - 1 x 2. The right section is y = 2 for 2 < x < 3. Combining these formulas, we write f ( x ) = - ( x + 2) if - 3 < x < - 1 x if - 1 x 2 2 if 2 < x < 3 . The graph of y = f ( x - 1) is that of f shifted to the right 1 unit. -1 -2 -3 1 2 3 4 0 -1 1 2 x y
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2. Functions 39 The graph of y = f ( x - 1) + 2 is that of y = f ( x - 1) shifted up 2 units. -1 -2 -3 1 2 3 4 0 -1 1 2 3 4 x y Example 2.30. Express the function f ( x ) = x + | x | as a piecewise- defined formula and sketch its graph. Solution: The graph consists of parts of two lines. For the part x < 0, we have that | x | = - x and so f ( x ) = x + | x | = x + ( - x ) = 0. For the right part, if x 0 we have that | x | = x , so f ( x ) = x + | x | = x + x = 2 x . Combining these formulas, we obtain f ( x ) = 0 if x < 0 2 x if x 0 . 0 x y
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40 J. S´ anchez-Ortega Example 2.31. Sketch the graph of the function y = | x 2 - 1 | . Give the piecewise formula. Solution: Let us take first graph the parabola y = x 2 - 1 by shifting the parabola y = x 2 downward 1 unit. -1 1 0 x y Notice that when - 1 < x < 1 we have that x 2 - 1 < 0, which yields that | x 2 - 1 | = - ( x 2 - 1) = 1 - x 2 . For x 1 or x ≤ - 1 we have that x 2 - 1 0, and so | x 2 - 1 | = x 2 - 1. Combining these formulas we get that f ( x ) = x 2 - 1 if x ≤ - 1 1 - x 2 if - 1 < x < 1 x 2 - 1 if x 1 To sketch the graph of y = | x 2 - 1 | we reflect the part of the parabola y = x 2 - 1 for - 1 < x < 1. We could also think about shifting the graph of y = - x 2 upward 1 unit. -1 1 0 x y
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2. Functions 41 2.11 Combinations of Functions Like numbers, functions can be added, subtracted, multiplied, and divided (except where the denominator in zero) to produce new functions. Definitions 2.32. If f and g are functions, then for every x that belongs to the domain of both f and g we define functions f + g , f - g , fg , f/g by the formulas: ( f + g )( x ) = f ( x ) + g ( x ) ( f - g )( x ) = f ( x ) - g ( x ) ( fg )( x ) = f ( x ) g ( x ) ( f/g )( x ) = f ( x ) /g ( x ) , where g ( x ) 6 = 0 . If the domain of f is A and the domain of g is B , then the domain of f + g is the intersection A B because both f ( x ) and g ( x ) have to be defined.
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