20 12 The equation of a curve is \( y=x^{4}-8 x^{2}+5 \). (a) Find the derivative, \( \left(\frac{d y}{d x}\right) \), of \( y=x^{4}-8 x^{2}+5 \). [2] (b) Find the coordinates of the three turning points. You must show all your working. \( \qquad \) and ( \( \qquad \) ) and ( \( \qquad \) [4] (c) Determine which one of these turning points is a maximum. Justify your answer.

Let's tackle your calculus problem step by step! (a) To find the derivative \( \frac{dy}{dx} \) of the equation \( y = x^4 - 8x^2 + 5 \), we differentiate term by term. Using the power rule: \[ \frac{dy}{dx} = \frac{d}{dx}(x^4) - \frac{d}{dx}(8x^2) + \frac{d}{dx}(5) = 4x^3 - 16x + 0 = 4x^3 - 16x. \] Thus, \( \frac{dy}{dx} = 4x^3 - 16x \). (b) To find the turning points, we need to set the derivative equal to zero: \[ 4x^3 - 16x = 0. \] Factoring out \(4x\): \[ 4x(x^2 - 4) = 0 \implies 4x(x - 2)(x + 2) = 0. \] So, the solutions are \(x = 0\), \(x = 2\), and \(x = -2\). Now we find the corresponding \(y\)-coordinates by substituting these \(x\) values back into the original equation: 1. For \(x = 0\): \[ y = 0^4 - 8(0)^2 + 5 = 5 \quad \Rightarrow (0, 5) \] 2. For \(x = 2\): \[ y = 2^4 - 8(2)^2 + 5 = 16 - 32 + 5 = -11 \quad \Rightarrow (2, -11) \] 3. For \(x = -2\): \[ y = (-2)^4 - 8(-2)^2 + 5 = 16 - 32 + 5 = -11 \quad \Rightarrow (-2, -11) \] Thus, the turning points are \( (0, 5) \), \( (2, -11) \), and \( (-2, -11) \). (c) To determine which turning point is a maximum, we will use the second derivative test. First, we find the second derivative: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(4x^3 - 16x) = 12x^2 - 16. \] Now, we evaluate the second derivative at each turning point: 1. At \(x = 0\): \[ \frac{d^2y}{dx^2} = 12(0)^2 - 16 = -16 < 0 \quad (\text{local maximum}) \] 2. At \(x = 2\): \[ \frac{d^2y}{dx^2} = 12(2)^2 - 16 = 48 - 16 = 32 > 0 \quad (\text{local minimum}) \] 3. At \(x = -2\): \[ \frac{d^2y}{dx^2} = 12(-2)^2 - 16 = 48 - 16 = 32 > 0 \quad (\text{local minimum}) \] Therefore, the turning point \( (0, 5) \) is a local maximum, confirmed by the second derivative being negative.