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Here is the solution to question 3. 3. Given the function f(x) = (x^2 - 1)/(x^2 + 1). a) Find f'(x). Step 1: Identify u and v for the quotient rule. Let u = x^2 - 1 and v = x^2 + 1. Then, find their derivatives: u' = 2x and v' = 2x. Step 2: Apply the quotient rule, f'(x) = (u'v - uv')/(v^2). f'(x) = ((2x)(x^2 + 1) - (x^2 - 1)(2x))/((x^2 + 1)^2) f'(x) = (2x^3 + 2x - (2x^3 - 2x))/((x^2 + 1)^2) f'(x) = (2x^3 + 2x - 2x^3 + 2x)/((x^2 + 1)^2) f'(x) = (4x)/((x^2 + 1)^2) The derivative is f'(x) = (4x)/((x^2 + 1)^2). b) Find the coordinates of the turning points of f(x). Step 1: Set f'(x) = 0 to find the critical points. (4x)/((x^2 + 1)^2) = 0 This implies 4x = 0, so x = 0. Step 2: Substitute the x-value back into f(x) to find the corresponding y-coordinate. f(0) = ((0)^2 - 1)/((0)^2 + 1) = (-1)/(1) = -1 The coordinate of the turning point is (0, -1). c) Determine the nature of the turning points. Step 1: Find the second derivative, f''(x). We use the quotient rule for f'(x) = (4x)/((x^2 + 1)^2). Let u = 4x u' = 4. Let v = (x^2 + 1)^2 v' = 2(x^2 + 1)(2x) = 4x(x^2 + 1). f''(x) = (4(x^2 + 1)^2 - (4x)(4x(x^2 + 1)))/(((x^2 + 1)^2)^2) f''(x) = (4(x^2 + 1)^2 - 16x^2(x^2 + 1))/((x^2 + 1)^4) Factor out 4(x^2 + 1) from the numerator: f''(x) = (4(x^2 + 1)[(x^2 + 1) - 4x^2])/((x^2 + 1)^4) f''(x) = (4(1 - 3x^2))/((x^2 + 1)^3) Step 2: Evaluate f''(x) at the critical point x=0. f''(0) = (4(1 - 3(0)^2))/(((0)^2 + 1)^3) = (4(1))/(1^3) = 4 Step 3: Determine the nature of the turning point. Since f''(0) = 4 > 0, the turning point (0, -1) is a local minimum. The nature of the turning point is a local minimum. 3 done, 2 left today. You're making progress.