Math 181 Spring 2007 HW 1 Corrected

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1 Math 181 Spring 2007 HW 1 Corrected February 1, 2007 Sec. 1.1 # 2 The graphs of f and g are given (see the graph in the book). (a) State the values of f( 4) and g(3). Find 4 on the x-axis (horizontal axis) move up or down to the red graph which is f the move over to the y-axis (vertical axis) to get f( 4) = 2. For g(3) proceed similarly except use the blue graph which is g. This gives g(3) = 4. (b) For what values of x is f(x) = g(x)? Whenever the graphs cross each other or intersect, f(x) = g(x). So from the graph we see this happens at x = 2 and x = 2. (c) Estimate the solution of f(x) = 1. Find 1 on the vertical axis then move over to the f graph, then up to the x-axis to get f(x) = 1 when x = 3. Similarly, we nd f(x) = 1 when x = 4. (d) On what interval is f decreasing? f is decreasing on an interval if f(x 1 ) > f(x 2 ) whenever x 1 < x 2 on the interval. In other words, the function is getting smaller as we move from left to right. This is happening for f on the interval 0 < x < 4. (e) State the domain and range of f. The domain consists of all numbers x that we can plug into the function and get a y value. For this function, from the picture (the red graph) we that that the domain is f x j 4 x 4 g. The range consists of everything we get when we plug everything from the domain into the function. In this case, from the picture (the red graph) we see that the range is f y j 2 y 3 g. (f) State the domain and range of g. The domain consists of all numbers x that we can plug into the function and get a y value. For this function, from the picture (the blue graph) we that that the domain is f x j 4 x 3 g. The range consists of everything we get when we plug everything from the domain into the function. In this case, from the picture (the blue graph) we see that the range is f y j 1 2 y 4 g. 1

2 Sec. 1.1 # 6 Determine whether the curve (pictured in the book) is the graph of a function of x. If it is, state the domain and range of the function. To be a function, each element in the domain must be associated (mapped, paired) with exactly one element in the range. This function passes that test (the vertical line test). Thus it is a function. As in the previous problem from the picture we see that the domain is f x j 2 x 2 g and the range is f y j 1 y 2 g. Sec. 1.1 # 22 Use the function f(x) = x f(x + h) f(x) and nd (a)f(2 + h), f(x + h), and, where h 6= 0. x + 1 h For f(2 + h) we replace each x in the denition of f with 2 + h and then simplify if possible. This gives f(2 + h) = (2 + h) (2 + h) + 1 = 2 + h 3 + h : You cannot cancel!!! If you want to, talk to me so that I can teach you why not. Your algebra will improve. For f(x + h) we replace each x in the denition of f with x + h and then simplify if possible. This gives f(x + h) = (x + h) (x + h) + 1 = x + h x + h + 1 : You cannot cancel!!! Again, if you want to, talk to me so that I can teach you why not. Your algebra will improve. For f(x + h) h f(x) ; replace f(x + h) with the last result, replace f(x) with its denition, and then simplify if possible. Thus f(x + h) h f(x) = x + h x + h + 1 h x x + 1 = x + h ( x + h + 1 h x x + 1 ) (x + 1) (x + 1) 2

3 = = (x + h)(x + 1) x + h + 1 h (x + 1) (x + h)(x + 1) ( x + h + 1 h (x + 1) x x) (x + h + 1) (x + h + 1) = (x + h) (x + 1) x (x + h + 1) h (x + 1)(x + h + 1) = x 2 + x h + x + h (x 2 + x h + x) h (x + 1)(x + h + 1) = x 2 + x h + x + h x 2 x h x h (x + 1)(x + h + 1) = = h h (x + 1)(x + h + 1) 1 (x + 1)(x + h + 1) : So nally f(x + h) f(x) h = 1 (x + 1)(x + h + 1) : In step (1) we substitute f(x + h) and f(x) as described. In step (2) we multiply top and bottom by x + 1. In step (3) we distribute and cancel where possible. In step (4) we multiply top and bottom by x + h + 1. In step (5) we distribute and cancel where possible. In step (6) we expand the terms in the numerator. In step (7) we distribute the minus sign or the 1. In step (8) we combine like terms, and in step (9) we cancel the h. You could do this in a dierent order and in slightly dierent ways and get the same result. Sec. 1.1 # 38 Find the domain and sketch the graph of the function f(x) = 2x + 3 if x < 1 ; 3 x if x 1 : Each portion of this denition is a straight line. The rst section which extends over the interval 1 < x < 1 has slope m = 2 and y intercept b = 3. The second section which extends over the interval 1 x < 1 has slope m = 1 and y intercept b = 3. We see from the denition that we can compute the value of the function for any real number thus the domain is f x j 1 < x 1 g. 3

4 We can use points on the rst line at x = 1 and at x = 3 to draw that portion of the graph, though the value of the line at x = 1 is not included in that portion of the graph (thus the open dot). At x = 1 we have 2( 1) + 3 = 1 so one ordered pair used for sketching is ( 1 ; 1 ). At x = 2 we have 2( 2) + 3 = 1 and our second ordered pair is ( 2 ; 1 ). Plotting these and sketching a line joining then gives the graph: Figure 1: y = 2 x + 3 for x < 1 : We can use points on the second line at x = 1 and at x = 3 to draw that portion of the graph, as the value of the line at x = 1 is included in that portion of the graph we have a solid dot. At x = 1 we have 3 ( 1) = 4 so one ordered pair used for sketching is ( 1 ; 4 ). At x = 3 we have 3 3 = 0 and our second ordered pair is ( 3 ; 0 ). Plotting these and sketching a line joining then gives the graph: Figure 2: y = 3 x for x 1 : 4

5 Putting both portions together gives the nished graph of the piecewise dened function. Figure 3: y = f(x) for 1 < x < 1 : Sec. 1.1 # 42 Find an expression for the function whose graph is the line segment joining the points ( 3 ; 2 ) and ( 6 ; 3 ). Find the slope rst using the formula Using our points we have m = m = y 1 y 2 x 1 x 2 : = 5 9 = 5 9 : Next using the point slope formula, y y 1 = m (x x 1 ), and using ( 6 ; 3 ) as our point, we have Adding 3 to both sides gives y 3 = 5 9 (x 6) = 5 9 x (5 9 ) 6 = 5 9 x 30 9 = 5 9 x 10 3 : Thus the function is f(x) = 5 9 x 1 3 y = 5 9 x = 5 9 x = 5 9 x 1 3 : with domain f x j 3 x 6 g. 5

6 Sec. 1.1 # 48 Find a formula for the desired function and state its domain. A rectangle has area 16 m 2. Express the perimeter of the rectangle as a function of the length of one of its sides. I will call the length of the rectangle x and the width y. Since the area is 16 m 2 and Area = x y we have x y = 16 : The equation for the perimeter, P is P = 2 x + 2 y. If we solve the area equation for either variable, say y, we have y = 16 x : We the substitute for y in the perimeter formula to get Thus P = 2 x + 2 y = 2 x + 2 ( 16 x ) = 2 x + 32 x : P = 2 x + 32 x : Now in order to have a rectangle, neither side can have 0 length, thus x must be greater than zero. Also the function is undened if x = 0. Also, the sides can't have negative length. x can be as big as we wish as this just requires that y get small to preserve the area requirement. Thus the domain of our perimeter function is f x j 0 < x < 1 g. Sec. 1.1 # 62 Determine whether f(x) = x 3 is even, odd, or neither. If f is even or odd, use symmetry to sketch its graph. First of all, recall that x 3 = 1 x. Next, a function is even if f( x) = f(x) and odd if 3 f( x) = f(x). Thus, let's compute f( x) and see what happens. f( x) = ( x) 3 = 1 ( x) = 1 3 (x 3 ) = 1 x = 3 (x 3 ) = f(x) : So, f is odd. If we sketch the positive or negative half of the function we can sketch the other half using the symmetry of an odd function. First mirror the half we sketch across the y axis and then mirror that half across the x axis. The positive half of the function looks similar to the graph of 1 = x except it increases faster as x gets close to 0 and it approaches the x axis faster as x gets large. The positive half graph is 6

7 Figure 4: y = x 3 for x 0 : If we reect this across the y axis and then across the x axis we get the following graph for the negative half of the function. Figure 5: y = x 3 for x 0; : 7

8 Putting both pieces together gives the nished graph below. Figure 6: y = x 3 for 1 < x < 1 : Sec. 1.2 # 2 Classify each function as a power function, root function, polynomial (state its degree), rational function, algebraic function, trigonometric function, exponential function, or logarithmic function. (a) y = x 6 x + 6 This is the ratio of two polynomials so it's a rational function and as it can be constructed by dividing two polynomials, it is algebraic. (b) y = x + x 2 p x 1 Because of the root this cannot be a polynomial or a rational function. It is not just a root, so it is not a root function. It can be constructed by starting with polynomials and performing a nite number of additions, subtractions, multiplications, divisions, and taking roots. Thus it is an Algebraic function. (c) y = 10 x This is an exponential function. The base is 10. It is also transcendental. (d) y = x 10 8

9 This is a power function, a polynomial, and algebraic. The degree is 10. (e) y = 2 t 6 + t 4 This is a polynomial of degree 6 and thus also algebraic.. Don't get confused by the constant. (f) y = cos + sin This is a trigonometric function and thus transcendental. Sec. 1.3 # 4 The graph of f is given. Draw the graphs of the following functions. Figure 7: f(x) (a) y = f(x + 4) This shifts the entire graph to the left by 4. Figure 8: f(x + 4) 9

10 (b) y = f(x) + 4 This shifts the entire graph up by 4. Figure 9: y = f(x) + 4 (c) y = 2 f(x) This stretches the graph vertically by a factor of two. I suggest you compute the corners, move them, and then connect the dots to get the new graph. The original y value for the far left corner is 0. If we multiply 0 by 2 we still have 0 so that point is unchanged. The next point has a y value of 4. If we multiply 4 by 2 we get 8 so our new point is ( 0 ; 8 ). The next point is ( 2 ; 1 ). Scaling the y value by 2 gives the new point ( 2 ; 2 ). The last point is ( 4 ; 2 ). Scaling by a factor of 2 gives the new point ( 4 ; 4 ). Connecting the dots gives the scaled graph. Figure 10: y = 2 f(x) 10

11 1 (d) y= f(x) + 3 The 1 compresses 2 2 f by a factor of 1. I would connect the dots as in the last 2 answer. Next the minus sign reects the new function across the xaxis, and nally, we add 3 to this function which moves the result up by 3. First the 1 factor. The leftmost point does not change as the y value is 0. The next point changes 2 from ( 0 ; 4 ) to ( 0 ; 2 ). The next point changes from ( 2 ; 1 ) to ( 2 ; 1 ), and the last point changes 2 from ( 4 ; 2 ) to ( 4 ; 1 ). We connect the dots to get... Figure 11: y = 1 2 f(x) Next we apply the minus sign to reect this result across the x axis to obtain: Figure 12: y = 1 2 f(x) 11

12 Lastly, we add 3 to this function to obtain our desired result. Figure 13: y = 1 2 f(x) + 3 Generally, if we wish to apply multiple shifts, scalings, and mirrorings, there is more than one order that will yield the same result. For example, in this case, we could have mirrored, scaled and then added to describe just one. Sec. 1.3 # 6 The graph of y = p 3 x x 2 is given. Use transformations to create the graph shown. The new graph is not scaled horizontally but it is scaled vertically by a factor of 2 and shifted to the right by 2 units. Thus if the old function is f(x), then the new function is 2 f(x 2). In our case as y = p 3 x x 2 the new function is y = 2 p 3 (x 2) (x 2) 2 = 2 p 3 x 6 (x 2 4 x + 4) = 2 p x x 10 : 12

13 Sec. 1.3 # 10 Graph y = 1 x 2 by starting with one of the standard graphs given in Section 1.2 and then applying the appropriate transformations. We start with the power function or polynomial y = x 2. Figure 14: y = x 2 Next, we multiply the function by 1 which reects the graph across the x axis and gives y = x 2. Figure 15: y = x 2 13

14 Finally, we add 1 to the function which moves the graph up 1 and gives y = x = 1 x 2. Figure 16: y = 1 x 2 Sec. 1.3 # 42 Find f g h where f(x) = 2 x 1 ; g(x) = x 2 ; and h(x) = 1 x. Well... f g h = f(g(h(x))) = f(g(1 x)) = f((1 x) 2 ) = f(1 2 x + x 2 ) = 2 (1 2 x + x 2 ) 1 = 2 4 x + 2 x 2 1 = 1 4 x + 2 x 2 : So f g h = 1 4 x + 2 x 2 : Sec. 1.3 # 46 Express the function F (x) = sin( p x ) in the form f g. We need to choose an f and a g so that f g = f(g(x)) = sin( p x ). There is in general more than one choice for the solution of these type of problems although in this case the choice should be let 14

15 f(x) = sin x and let g(x) = p x. Then f g = f(g(x)) = f( p x ) = sin( p x ) ; as desired. 15

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