Find all open intervals on which the function \( f(x)=x^{4}-2 x^{2}+3 \) is increasing and decreasing. Then find all open stervals on which this function is concave up and concave down.

To analyze the function \( f(x) = x^4 - 2x^2 + 3 \), start by finding its first derivative, which helps determine where the function is increasing or decreasing. The first derivative is: \[ f'(x) = 4x^3 - 4x = 4x(x^2 - 1) = 4x(x - 1)(x + 1). \] Setting \( f'(x) = 0 \) gives critical points: \[ x = -1, 0, 1. \] To determine where \( f'(x) \) is positive (increasing) or negative (decreasing), test intervals around the critical points: 1. For \( x < -1 \) (e.g., \( x = -2 \)): \( f'(-2) = 4(-2)(-1)(-3) < 0 \) (decreasing). 2. For \( -1 < x < 0 \) (e.g., \( x = -0.5 \)): \( f'(-0.5) = 4(-0.5)(-1.5)(0.5) > 0 \) (increasing). 3. For \( 0 < x < 1 \) (e.g., \( x = 0.5 \)): \( f'(0.5) = 4(0.5)(-0.5)(1.5) < 0 \) (decreasing). 4. For \( x > 1 \) (e.g., \( x = 2 \)): \( f'(2) = 4(2)(1)(3) > 0 \) (increasing). Thus, the function is increasing on the intervals: \[ (-1, 0) \cup (1, \infty) \] and decreasing on the interval: \[ (-\infty, -1) \cup (0, 1). \] Next, find the second derivative to analyze concavity: \[ f''(x) = 12x^2 - 4. \] Set \( f''(x) = 0 \): \[ 12x^2 - 4 = 0 \Rightarrow 12x^2 = 4 \Rightarrow x^2 = \frac{1}{3} \Rightarrow x = \pm\frac{1}{\sqrt{3}}. \] Test intervals for concavity: 1. For \( x < -\frac{1}{\sqrt{3}} \) (e.g., \( x = -2 \)): \( f''(-2) = 12(-2)^2 - 4 > 0 \) (concave up). 2. For \( -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \) (e.g., \( x = 0 \)): \( f''(0) = -4 < 0 \) (concave down). 3. For \( x > \frac{1}{\sqrt{3}} \) (e.g., \( x = 2 \)): \( f''(2) > 0 \) (concave up). This leads us to the intervals: Concave up on: \[ (-\infty, -\frac{1}{\sqrt{3}}) \cup (\frac{1}{\sqrt{3}}, \infty) \] and concave down on: \[ (-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}). \]