6. Given the function \( \quad y=x^{4}-2 x^{2}-2 \) Determine all possible max, min points and inflection points

To find the maximum and minimum points of the function \( y = x^4 - 2x^2 - 2 \), we start by finding the first derivative: \[ y' = 4x^3 - 4x = 4x(x^2 - 1) = 4x(x - 1)(x + 1) \] Setting \( y' = 0 \) gives us the critical points: \[ x = 0, \quad x = 1, \quad x = -1 \] Next, we use the second derivative to determine concavity and the nature of these critical points: \[ y'' = 12x^2 - 4 \] Evaluating the second derivative at each critical point: 1. At \( x = 0 \): \[ y''(0) = 12(0)^2 - 4 = -4 \quad \text{(concave down, local max)} \] 2. At \( x = 1 \): \[ y''(1) = 12(1)^2 - 4 = 8 \quad \text{(concave up, local min)} \] 3. At \( x = -1 \): \[ y''(-1) = 12(-1)^2 - 4 = 8 \quad \text{(concave up, local min)} \] Now we can summarize that there is a local maximum at \( (0, -2) \) and local minima at \( (1, -3) \) and \( (-1, -3) \). For inflection points, we want to set the second derivative equal to zero: \[ 12x^2 - 4 = 0 \quad \Rightarrow \quad x^2 = \frac{1}{3} \quad \Rightarrow \quad x = \pm \frac{\sqrt{3}}{3} \] Evaluating the function at these points gives us the inflection points \( \left( \frac{\sqrt{3}}{3}, y\left( \frac{\sqrt{3}}{3} \right) \right) \) and \( \left( -\frac{\sqrt{3}}{3}, y\left( -\frac{\sqrt{3}}{3} \right) \right) \), where you can plug \( x \) back into the original function to find the corresponding \( y \)-values.