course MTH 271 12/15,1200 assignment #028028. `query 28
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12:04:59 Query 3.7.12 sketch y = -x^3+3x^2+9x-2 **** Describe your graph, including a description of any intercepts, extrema, asymptotes and concavity.
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RESPONSE --> Graph is a cubic curve, where a < 0 x-intercepts:-2, .209, 4.69 No horizontal/vertical asymptotes (-infinity,-1):decreasing (-1,3):increasing (3,infinity):decreasing (-1,-5): Relative Minimum (3,25): Relative Maximum (1, 9):Point of Inflection (-infinity,1):concave upward (1,infinity): concave downward confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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12:05:13 First we find the zeros: You can find the zero at x = -2 by inspection (i.e., try a few simple values of x and see if you 'hit' one). Knowing that x = -2 is a zero, you know that (x - (-2) ) = x + 2 is a factor of the expression. You can find the other factor by long division of x + 2 into -x^3 + 3 x^2 + 9x - 2. You get quotient x^2 - 5 x + 1. Thus -x^3 + 2 x^2 + 9 x - 2 = - (x + 2)·(x^2 - 5·x + 1). This expression is zero when x+2 = 0 or when x^2 - 5 x + 1 = 0. The first equation we already know gives us x = -2. The second is solved by the quadratic formula. We get x = [ -(-5) +- `sqrt( (-5)^2 - 4 * 1 * 1 ) ] / (2 * 1) = [ 5 +- `sqrt(21) ] / 2. Simplifying we get approximate x values .21 and 4.7. Then we find maxima and minima using 1st and 2d derivative: The derivative is -3x^2+6x + 9, which gives the equation -3x^2+6x + 9 = 0 for critical points. Dividing thru by -3 we get x^2 - 2x - 3 = 0 or(x-3)(x+1) = 0 so x = 3 or x = -1. Second derivative is -6x + 6, which is negative when x = 3 and positive when x = -1. At x = 3 we have a maximum. Evaluating y = -x^3+3x^2+9x-2 at x = 3 we get y = 25. The negative second derivative indicates that (3,25) is a maximum. At x = -1 we have a minimum. Evaluating y = -x^3+3x^2+9x-2 at x = -1 we get y = -7. The positive second derivative indicates that this (-1,-7) is a minimum. Finally we analyze 2d derivative for concavity and pts of inflection: The second derivative -6x + 6 is zero when x = 1; at this point the derivative, which is linear in x, changes from positive to negative. Thus the x = 1 point (1, 9) is a point of inflection. The derivative is positive and the function therefore concave up on (-infinity, 1). The derivative is negative and the function therefore concave down on (1, infinity). The function is defined for all x so there are no vertical asymptotes. As | x | -> infinity the magnitude of the function -> infinity so there are no horizontal asymptotes. **
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3
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12:08:45 Query 3.7.34 sketch (x^2+1)/(x^2-1) Note: The problem in the text might be (x^2+1)/(x^2-2). If so the solution given below can be easily adapted to that function. Describe your graph, including a description of any intercepts, extrema, asymptotes and concavity.
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RESPONSE --> Vertical Asymptotes at x=+-1 Horizontal Asymptote at y=a/b: y=1/1 =1 (-infinity,-1)=concave upward (-1,1)=concave downward (1,infinity)=concave upward confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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12:08:57 First we look for zeros and intercepts: The numerator is never zero, being the sum of the positive number 1 and the nonnegative quantity x^2. So the function has no zeros, i.e., never crosses the x axis. When x=0 we have y = (0^2+1)/(0^2-1) = -1 so the y intercept is (0,-1). Next we analyze the derivative to see if we can find relative maxima and minima: The derivative is - 4x/(x^2 - 1)^2, which has its only zero when x = 0. So (0, -1) is the only critical point. The second derivative is 4(3x^2 + 1)/(x^2 - 1)^3, which is negative when x = 0. So the critical point gives us a maximum. We analyze the second derivative to determine concavity: The second derivative 4(3x^2 + 1)/(x^2 - 1)^3 has a numerator which is always positive, since x^2 is always positive. The denominator is negative where x^2 - 1 < 0, which occurs between x = -1 and x = 1. So the 2d derivative is positive on (-infinity, -1) and on (1, infinity), where the graph will be concave up, and negative on (-1, 1), where the graph will be concave down. Now we look for vertical and horizontal asymptotes: The denominator of (x^2+1)/(x^2-1) is zero and the numerator isn't when x = +1 and also when x = -1. So we have vertical asymptotes at x = +1 and at x = -1. }For very large positive x or for very large negative x the +1 in the numerator and the -1 in the denominator are both insignificant and the function value is very close to x^2 / x^2 = 1. The function approaches its horizontal asymptote y = 1 for both large positive x and large negative x. We finally determine where the function is positive and where negative: For x < -1 the function cannot change sign, since it is continuous and has no zeros. Testing any x < -1 gives us a positive value. On this interval the function is therefore positive. The same is true for x > 1. For -1 < x < 1 the same argument shows that the function is negative**
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RESPONSE --> ------------------------------------------------ Self-critique Rating:ent: 3"