course Mth 158 4/12 7:45 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * If the farmer uses 2000 meters of fence, and fences x feet along the highway, then he have 2000 - x meters for the other two sides. So the dimensions of the rectangle are x meters by (2000 - x) / 2 meters. The area is therefore x * (2000 - x) / 2 = -x^2 / 2 + 1000 x. The graph of this function forms a downward-opening parabola with vertex at x = -b / (2 a) = -1000 / (2 * -1/2) = 1000. At x = 1000 the area is -x^2 / 2 + 1000 x = -1000^2 / 2 + 1000 * 1000 = 500,000, meaning 500,000 square meters. Since this is the 'highest' point of the area vs. dimension parabola, this is the maximum possible area. ** STUDENT QUESTION: I got the formula for the area but don't understand how you get a parabola out of that. INSTRUCTOR RESPONSE: There are two overall steps to solving this problem. The first is to get an expression for the area. The second is to find the maximum possible value of the expression. You have the first step, which gives you the equation A = 1000 x - x^2 / 2. The second step is to find the maximum of your expression. To find the maximum of the expression 1000 x - x^2 / 2, you can consider the graph of the function. The expression can be written as - x^2 / 2 + 1000 x. This is in the form of a quadratic expression of form a x^2 + b x + c with a = -1/2, b = 1000 and c = 0. As you know from your previous work a quadratic function has a graph which is a parabola, and the vertex of a parabola is either its highest or lowest point (depending on whether it opens downward or upward). The vertex of the parabola occurs when x = - b / (2 a). Substituting b = 1000 and a = -1/2 we find that x_vertex = - 1000 / (2 * (-1/2) ) = 1000. The corresponding y coordinate is y = -1/2 x_vertex^2 + 1000 * x_vertex = -1/2 * 1000^2 + 1000 * 1000 = 500 000. Since a = -1/2, which is negative, the parabola opens downward. This makes the vertex the highest point, so the value of the function is maximized at the vertex. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: * 4.4.33 / 7th edition 4.1.101 (was 4.1.80). A rectangle has one vertex on the line y=10-x, x>0, another at the origin, one on the positive x-axis, and one on the positive y-axis. Find the largest area that can be enclosed by the rectangle. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (10-x) * x = 10x-x^2 = -(10)/2(-1) = -10/-2 = 5 Vertex = 5 -x^2 + 10x = -(5)^2 + 10 ( 5) = 25 confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The dimensions of the rectangle are x and y = 10 - x. So the area is area = x ( 10 - x) = -x^2 + 10 x. The vertex of this rectangle is at x = -b / (2 a) = -10 / (2 * -1) = 5. Since the parabola opens downward this value of x results in a maximum area, which is -x^2 + 10 x = -5^2 + 10 * 5 = 25. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*2 "
course Mth 158 4/12 8:40 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * This is not a polynomial function. It is the ratio of two polynomials, the ratio of x^2 - 5 to x^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): the ratio of x^2 - 5 to x^3. ------------------------------------------------ Self-critique rating #$&*2 ********************************************* Question: * 5.1.20 / 7th edition 4.2.40 (was 4.2.30). If a polynomial has zeros at x = -4, 0, 2 the what is its minimum possible degree, and what form will the polynomial have? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [x – (-4)][(x-(0)][x-(2)]= (x+4)(x-0)(x-2) confidence rating #$&*1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The factors of the polynomial must include (x - (-4) ) = x + 4, x - 0 = x, and x - 2. So the polynomial must be a multiple of (x+4)(x)(x-2). The general form of the polynomial is therefore f(x)=a(x+4)(x-0)(x-2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): f(x)=a(x+4)(x-0)(x-2). ------------------------------------------------ Self-critique rating #$&*2 ********************************************* Question: * 5.1.42 / 7th edition 4.2.52 (was 4.2.40). What are the zeros of the function f(x)=(x+sqrt(3))^2 (x-2)^4 and what is the multiplicity of each? Your solution [(x+sqrt(3)]=0 X= - sqrt(3) with 2nd degree (x-2)=0 X=2 with fourth degree confidence rating #$&*0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * f(x) will be zero if x + sqrt(3) = 0 or if x - 2 = 0. The solutions to these equations are x = - sqrt(3) and x = 2. The zero at x = -sqrt(3) comes from (x + sqrt(3))^2 so has degree 2. The zero at x = 2 comes from (x-2)^4 so has degree 4. For each zero does the graph touch or cross the x axis? In each case the zero is of even degree, so it just touches the x axis. Near x = -sqrt(3) the graph is nearly a constant multiple of (x+sqrt(3))^2. Near x = 2 the graph is nearly a constant multiple of (x - 2)^4. What power function does the graph of f resemble for large values of | x | ? If you multiply out all the terms you will be a polynomial with x^6 as the highest-power term, i.e., the 'leading term'. For large | x | the polynomial resembles this 'leading term'--i.e., it resembles the power function x^6. ** STUDENT QUESTION: I didn’t see the question of the power function until now. I am still a little unsure how to solve this. I have been reading ch5 again and the way I understand the power function at large values of |x| is this: There are only 4 types. 1. It opens up like that of a parabola when the power function is an even number (2, 4, 6….) and the leading number is greater than 0. 2. It open down when the power function is even and the leading number is less than 0. 3. It looks like an asymmetric graph with respect to the origin with the points (1,1) and (-1,-1) when the power function is odd and the leading number is greater than 0. 4. It looks like number 3 only it has been flipped with points (-1, -1) and (1, -1) when the power function is odd and the leading number is less than 0. If my interpretation is correct, then how do you multiply out the terms to get the highest power term? It seems like I have more trouble with the basics like (multiplying out the terms, and long division of polynomials) than grasping this. INSTRUCTOR RESPONSE Your list of the behaviors of the power functions is correct, as well expressed. Your list applies only to positive integer powers, That is appropriate here because we are dealing with polynomial functions, whose zeroes involve only positive integer powers. The function given in this problem involves even powers of both factors. To illustrate both even and odd powers, we're going to analyze here the function h(x) = (x+sqrt(3))^2 (x-2)^3 rather than the given function f(x)=(x+sqrt(3))^2 (x-2)^4: h(x)=(x+sqrt(3))^2 (x-2)^3 has zeroes when x + sqrt(3) = 0 and when x - 2 = 0. • The zeroes are therefore at x = - sqrt(3) and x = 2. • The zero at x = - sqrt(3) has multiplicity 2, and the zero at x = 2 has multiplicity 3. • Near x = - sqrt(3) the graph therefore has the same basic shape as the power function y = (x + sqrt(3)) ^2, with vertex at x = - sqrt(3). Since at x = -sqrt(3) the other factor (x - 2)^3 is negative, the graph will open downward. • Near x = 2 the graph will have the same basic shape as the power function y = (x - 2)^3, passing through the x axis at (0, 2). The graph of h(x) is depicted in the figure below: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Large value = 2 +4 = 6 ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: * 5.1.61 / 7th edition 4.2.62 (was 4.2.50). f(x)= 5x(x-1)^3. Give the zeros, the multiplicity of each, the behavior of the function near each zero and the large-|x| behavior of the function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5x=0 X=0 with degree of 1 x-1=0 x=0 with degree of 3 large value = 4 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The zeros occur when x = 0 and when x - 1 = 0, so the zeros are at x = 0 (multiplicity 1) and x = 1 (multiplicity 3). Each zero is of odd degree so the graph crosses the x axis at each. If you multiply out all the terms you will be a polynomial with 5 x^4 as the highest-power term, i.e., the 'leading term'. For large | x | the polynomial resembles this 'leading term'--i.e., it resembles the power function 5 x^4. What is the maximum number of turning points on the graph of f? This is a polynomial of degree 4. A polynomial of degree n can have as many as n - 1 turning points. So this polynomial could possibly have as many as 4 - 1 = 3 turning points. Give the intervals on which the graph of f is above and below the x-axis this polynomial has zeros at x = 0 and x = 1. So on each of the intervals (-infinity, 0), (0, 1) and (1, infinity) the polynomial will lie either wholly above or wholly below the x axis. If x is a very large negative or positive number this fourth-degree polynomial will be positive, so on (-infinity, 0) and (1, infinity) the graph lies above the x axis. On (0, 1) we can test any point in this interval. Testing x = .5 we find that 5x ( x-1)^3 = 5 * .5 ( .5 - 1)^3 = -.00625, which is negative. So the graph lies below the x axis on the interval (0, 1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Each zero is of odd degree so the graph crosses the x axis at each. This is a polynomial of degree 4. A polynomial of degree n can have as many as n - 1 turning points. So this polynomial could possibly have as many as 4 - 1 = 3 turning points. Give the intervals on which the graph of f is above and below the x-axis this polynomial has zeros at x = 0 and x = 1. So on each of the intervals (-infinity, 0), (0, 1) and (1, infinity) the polynomial will lie either wholly above or wholly below the x axis. If x is a very large negative or positive number this fourth-degree polynomial will be positive, so on (-infinity, 0) and (1, infinity) the graph lies above the x axis. On (0, 1) we can test any point in this interval. Testing x = .5 we find that 5x ( x-1)^3 = 5 * .5 ( .5 - 1)^3 = -.00625, which is negative. So the graph lies below the x axis on the interval (0, 1). ------------------------------------------------ Self-critique rating #$&*2 "