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 solve the problem step by step. ### Part (a): Find the derivative \( \frac{dy}{dx} \) The given equation is: \[ y = x^4 - 8x^2 + 5 \] To find the derivative, we will differentiate each term with respect to \( x \): 1. The derivative of \( x^4 \) is \( 4x^3 \). 2. The derivative of \( -8x^2 \) is \( -16x \). 3. The derivative of the constant \( 5 \) is \( 0 \). Putting it all together, we have: \[ \frac{dy}{dx} = 4x^3 - 16x \] ### Part (b): Find the coordinates of the three turning points Turning points occur where the derivative is equal to zero. We set the derivative equal to zero: \[ 4x^3 - 16x = 0 \] Factoring out the common term: \[ 4x(x^2 - 4) = 0 \] This gives us: \[ 4x = 0 \quad \text{or} \quad x^2 - 4 = 0 \] From \( 4x = 0 \), we get: \[ x = 0 \] From \( x^2 - 4 = 0 \), we get: \[ x^2 = 4 \implies x = 2 \quad \text{or} \quad x = -2 \] Thus, the \( x \)-coordinates of the turning points are \( x = -2, 0, 2 \). Next, we find the corresponding \( y \)-coordinates by substituting these \( x \)-values back into the original equation: 1. For \( x = -2 \): \[ y = (-2)^4 - 8(-2)^2 + 5 = 16 - 32 + 5 = -11 \] 2. For \( x = 0 \): \[ y = 0^4 - 8(0)^2 + 5 = 5 \] 3. For \( x = 2 \): \[ y = (2)^4 - 8(2)^2 + 5 = 16 - 32 + 5 = -11 \] Thus, the coordinates of the turning points are: - \( (-2, -11) \) - \( (0, 5) \) - \( (2, -11) \) ### Part (c): Determine which one of these turning points is a maximum To determine whether each turning point is a maximum or minimum, we can use the second derivative test. We first find the second derivative \( \frac{d^2y}{dx^2} \). Starting from the first derivative: \[ \frac{dy}{dx} = 4x^3 - 16x \] Now, we differentiate again: 1. The derivative of \( 4x^3 \) is \( 12x^2 \). 2. The derivative of \( -16x \) is \( -16 \). Thus, the second derivative is: \[ \frac{d^2y}{dx^2} = 12x^2 - 16 \] Now we evaluate the second derivative at each turning point: 1. For \( x = -2 \): \[ \frac{d^2y}{dx^2} = 12(-2)^2 - 16 = 12(4) - 16 = 48 - 16 = 32 \quad (\text{positive, so minimum}) \] 2. For \( x = 0 \): \[ \frac{d^2y}{dx^2} = 12(0)^2 - 16 = 0 - 16 = -16 \quad (\text{negative, so maximum}) \] 3. For \( x = 2 \): \[ \frac{d^2y}{dx^2} = 12(2)^2 - 16 = 12(4) - 16 = 48 - 16 = 32 \quad (\text{positive, so minimum}) \] ### Conclusion The turning point that is a maximum is at: \[ (0, 5) \] This is justified because the second derivative at this point is negative, indicating a local maximum.