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Assignment 2 Solutions.pdf

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Department
Mathematics
Course Code
Mathematics 0110A/B
Professor
Chris Brandl

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MATH 127 Fall 2012 Assignment 2 Solutions 1. For each part, sketch all of the given functions on the same set of axes. p p p (a) (i) f(x) = x (ii) g(x) = ▯ ▯6x (iii) h(x) = ▯2 2 ▯ x ▯ 1 5 4 3 y=f(x) 2 1 -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 y=g(x)-2 y=h(x) -3 -4 -5 1 (b) (i) f(x) = e ▯x (ii) g(x) = ▯e ▯x+1 (iii) h(x) = + e e ▯x 5 y=h(x) 4 3 2 y=f(x) 1 -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -2 y=g(x) -3 -4 -5 1 1 4 (c) (i) f(x) = (ii) g(x) = + 1 (iii) h(x) = x x + 1 x 5 y=g(x) 4 y=h(x) 3 2 1 y=f(x) -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -2 -3 -4 -5 1 2. Sketch each function. (a) f(x) = jx+1j+jx▯1j (hint: this can be expressed as a piecewise function, show the work in getting this piecewise function) We have that x + 1 ▯ 0 when x ▯ ▯1 and that x ▯ 1 ▯ 0 when x ▯ 1. We now consider the three intervals x < ▯1, ▯1 ▯ x < 1, and x ▯ 1. For the ▯rst interval x < ▯1 we have that x+1 < 0 and x▯1 < 0 so we get that f(x) = jx + 1j + jx ▯ 1j = ▯(x + 1) ▯ (x ▯ 1) = ▯x ▯ 1 ▯ x + 1 = ▯2x. In the second interval ▯1 ▯ x < 1 we have that x + 1 ▯ 0 and x ▯ 1 < 0 so we get that f(x) = jx + 1j + jx ▯ 1j = (x + 1) ▯ (x ▯ 1) = x + 1 ▯ x + 1 = 2. In the third interval x ▯ 1 we have that x + 1 ▯ 0 and x ▯ 1 ▯ 0 so we get that f(x) = jx + 1j + jx ▯ 1j = (x + 1) + (x ▯ 1) = x + 1 + x ▯ 1 = 2x. 8 > > ▯2x if x < ▯1 < Thus we have that f(x) = jx + 1j + jx ▯ 1j = 2 if ▯ 1 ▯ x < 1 > > : 2x if x ▯ 1 4 3.5 3 2.5 2 1.5 1 0.5 -4 -3 -2 -1 0 1 2 3 4 -0.5 -1 1 (b) f(x) = (hint: ▯rst look at the graph of x ▯ 4x + 6) x ▯ 4x + 6 2 First consider the graph of y = x ▯ 4x + 6. Completing the square we can re- write this as y = (x ▯ 2) + 2 so we see that this is a parabola shifted 2 to the right and 2 up. The graph of y = (x ▯ 2) + 1 goes through the point (2;2) and approaches +1 for large negative and positive values of x. From this we see that 2 1 f(x) = goes through the point (2;1) and y approaches 0 for large x ▯ 4x + 6 2 positive and1.5gative values of x. 1.25 1 0.75 0.5 0.25 -5 -4 -3-2 -1 0 1 2 3 4 5 -0.25 -0.5 -0.75 -1 3. For each pair of functions, ▯nd f▯g and g▯f and state the domain of each composition. 1 p (a) f(x) = , g(x) = x x ▯ 2 p1 We have that (f ▯g)(x) = x▯2. For x to be in the domain of f ▯g we must have p p that x ▯ 2 6= 0 (x 6= ▯4) AND we need x to be de▯ned (x ▯ 0). Thus, the domain of f ▯ g is [0;4) [ (4;1). r 1 We have that (g ▯ f)(x) = . For x to be in the domain of g ▯ f we must x ▯ 2 have that 1 ▯ 0 AND we need 1 to be de▯ned. The ▯rst requirement gives x▯2 x▯2 that x ▯ 2 and the second requirement gives that x 6= 2. Thus, the domain of g ▯ f is all x > 2. p p (b) f(x) = x + 2, g(x) = x ▯ 4 p p We have that (f ▯ g)(x) = x ▯ 4 + 2. For x to be in the domain of f ▯ g we p p must have that x ▯ 4+2 ▯ 0 AND that x ▯ 4 is de▯ned. The ▯rst requirement is always true and the second requirement gives x ▯ 4. Thus, the domain is all x ▯ 4. p p We have that (g ▯ f)(x) = x + 2 ▯ 4. For x to be in the domain of g ▯ f p p we must have that x + 2 ▯ 4 ▯ 0 AND we must have that x + 2 is de▯ned. The ▯rst requirement gives that x ▯ 14 and the second requirement gives that x ▯ ▯2. Thus, the domain is all x ▯ 14. 3 4. Give a possible equation of each graph. From observing the graph state whether the function is even, odd, or neither. (i) (ii) 7 3 6 5 2.5 2 4 3 1.5 1 2 1 0.5 -7 -6 -5 -4 -3 -2 --1 1 2 3 4 5 6 7 -5 -4 -3 -2 -1-0.5 1 2 3 4 5 -2 -1 -3 -1.5
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