Question+45+-+Dan+and+Kevin

__A)__

F(x) = x²/x²-1 lim F(x) = (infinity)² / (infinity)² x à infinity lim F(x) = 1 x à infinity lim F(x) = (infinity)² / (infinity)² x à infinity lim F(x) = 1 x à -infinity - Horizontal Asymptote at F(x) = 1 as x à +/- infinity F(x) = x² / x²-1 F(x) = x² / (x+1)(x-1) - Vertical Asymptotes at x = {1, -1}

__B)__

F(x) = x^2/(x^2-1) Find first derivative: F '(x) = 2x(x^2-1) - 2x(x^2) / (x^2-1)^2 F '(x) = 2x^3 - 2x - 2x^3 / (x^4 - 2x^2 + 1) F '(x) = -2x / (x+1)(x-1)(x+1)(x-1) Find critical points, where first derivative is zero or undefined: 0= -2x x=0 where first derivative is 0 0=x+1 0=x-1 x=1,-1 where first derivative is undefined Critical Values X = {1, 0, -1} For values less than x=-1, the first derivative is positive so the function is increasing from (-infinity, -1) For values between x=-1 and x=0, the first derivative is also positive, so the function is also increasing from (-1,0) Increasing = (-infinity, -1) (-1,0) For x values beween 0 and 1, and greater than 1, the first derivative is negative, so the function is decreasing. Decreasing= (0, 1) (1, infinity)

__C)__ Search through critical points to find local extrema: F(-1) = -2(-1) / (0)(-2)(0)(-2) F(-1) = 2 / 0 = undefined --> critical value but not local extrema F(1) = -2(1) / (2)(0)(2)(0) F(1) = -2 / 0 = undefined --> critical value but not local extrema F(0) = -2(0) / (1)(-1)(1)(-1) F(0) = 0 / 1 = 0 --> critical value and local maximum - Local Max F(0) = 0 - No local minumums __D)__ first derivative= F'(x) = -2x / (x^2-1)^2 second derivative= F"(x) = (x^2-1)^2(-2) - (-2x)(2(x^2-1)(2x) / (x^2-1)^4 F"(x) = -2(x^2-1)^2 + 8x^2(x^2-1) / (x^2-1)^4 F"(x) = 2(x^2-1)[3x^2+1] / (x^2-1)^4

F"(x) = 0 at x = [1, -1) For values less than x=-1, the second derivative is positive so the function is concaving upwards from (-infinity, -1), same with (1,infinity) CU = (-infinity, -1) (1, infinity) For x values between -1 and 1, the second derivative is negative so the function is concaving downwards from (-1,1) CD = (-1,1) The second derivative changes signs at x=0; therefore, x=0 is an inflection point. Inflection point: x=0

__E)__ Graph F(x) = x^2/x^2-1

Answer: C) -5/14

d/dx[y^3x + y^2x^2] = d/dx[6]

y^3(1) + 3y^2(d/dx)x + y^2(2x) + 2y(d/dx)x^2 = 0

(d/dx)[3y^2x + 2yx^2] = -y^3 - 2xy^2

d/dx = [ -y^3 - 2xy^2] / [3y^2x + 2yx^2]

d/dx = [ -(1)^3 - 2(2)(1^2)] / [3(1)^2(2) + 2(1)(2^2)]

d/dx = -5/14 - Answer C

-Dan G