This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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You're on a roll — Here's how to determine the nature of the stationary points for y = x^4 - 16x^2. Step 1: Find the first derivative, (dy)/(dx). (dy)/(dx) = (d)/(dx)(x^4 - 16x^2) = 4x^3 - 32x Step 2: Find the x-coordinates of the stationary points by setting the first derivative to zero. 4x^3 - 32x = 0 Factor out 4x: 4x(x^2 - 8) = 0 This gives three possible values for x: 4x = 0 x = 0 x^2 - 8 = 0 x^2 = 8 x = ±sqrt(8) = ± 2sqrt(2) The stationary points occur at x = 0, x = 2sqrt(2), and x = -2sqrt(2). Step 3: Find the second derivative, (d^2y)/(dx^2). (d^2y)/(dx^2) = (d)/(dx)(4x^3 - 32x) = 12x^2 - 32 Step 4: Determine the nature of the stationary points using the second derivative test. For x = 0: Substitute x=0 into the second derivative: (d^2y)/(dx^2) |_x=0 = 12(0)^2 - 32 = -32 Since -32 < 0, the stationary point at x = 0 is a local maximum. For x = 2sqrt(2): Substitute x=2sqrt(2) into the second derivative: (d^2y)/(dx^2) |_x=2sqrt(2) = 12(2sqrt(2))^2 - 32 = 12(4 × 2) - 32 = 12(8) - 32 = 96 - 32 = 64 Since 64 > 0, the stationary point at x = 2sqrt(2) is a local minimum. For x = -2sqrt(2): Substitute x=-2sqrt(2) into the second derivative: (d^2y)/(dx^2) |_x=-2sqrt(2) = 12(-2sqrt(2))^2 - 32 = 12(4 × 2) - 32 = 12(8) - 32 = 96 - 32 = 64 Since 64 > 0, the stationary point at x = -2sqrt(2) is a local minimum. Send me the next one 📸