Question

Use the guidelines to sketch the curve y = 2x^2/x^2 - 1. The domain is {x | x^2 - 1 0} = {x | x plusminus 1} The x- and y-intercepts are both 0. Since f(-x) = f(x), the function f is. The curve is symmetric about the y-axis. Since the denominator is 0 when x = plusminus1, we compute the following limits: Therefore the lines x = 1 and x = are vertical asymptotes. This information about limits and asymptotes enables us to draw the asymptotes in the top figure. Since f'(x) > 0 when x < 0 (x -1) and f '(x) < 0 when x > 0 (x 1), f is increasing on the intervals (-infinity, -1) and (-1,0) and decreasing on the intervals (0, 1) and (1, infinity). The only critical number is x = 0. Since f' changes from positive to negative at, f = 0 ] is a local maximum by the First Derivative Test.

Answer 1

"A) Domain: - { 2/x notequalto plusminus 1 } = (-infinity, -1): (-1, 1): (1, infinity) E) F'(x) = (x^2 - _4x - 2x^2(2x)/(x^2 - 1)^2 = 4x^3 - 4x - 4x^3/(x^2 - 1)^2 = -4x/(x^2 - 1)^2 f'(x) = -4x/(x^2 - 1)^2 G) f''(x) = (x^2 - 1)^2 (-4) + (4x) (2) (x^2 - 1) (2x)/(x^2 - 1)^4 = -4(x^4 + 1 - 2x^2) + 16x^2(x^2 - 1)/(x^2 - 1)^4 = -4x^4 - 4 + 8x^2 + 16x^4 - 16x^2/(x^2 - 1)^4 = 12x^4 - 8x^2 - 4/(x^2 - 1)^4 = -4(x^2 - 1) + 16x^2/(x^2-1)^3 = 12x^2 + 4/(x^2 - 1)^3 F''(x) = x^2 +/(x^2 - 1)^3 Since the simplified numerator, 12x^2 + 4 > 0 for all x we have f''(x) > 0 implies x^2 - 1 > 0 implies