#$&* course MTH 271 5/7/126:40pm 028. `query 28
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery problem 1 g 8th edition 3.7.34 7th edition 3.7.32 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The numerator has no zeros, so this graph never crosses the x-axis. When x =0, y = -1, and so this is where the graph crosses the y-axis. The denominator only is zero when x equals either 1 or -1, so these are vertical asymptotes. When x > 1, the graph tests positive, and when x < -1, the graph tests positive, while on the interval (-1, 1) the graph tests negative. These functions are continuous, so they cannot change their sign, making them positive on the whole intervals above. The first derivative should give critical points: -4x/(x^2 - 1)^2 → the only zero and critical point is at (0,0) and the graph is undefined at x = 1, -1. The second derivative: 4(3x^2 + 1)/(x^2 - 1)^3. This graph is positive on the intervals (-inf., -1) and (1, inf.). It is negative on the interval (-1, 1). This shows that the original graph is concave upwards on the intervals where the 2nd der. is positive and concave downwards on the interval where the 2nd der. is negative. Also, this graph is negative when x = 0, so the critical point from the first der. is a maximum. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!