Find the intervals on which \( f(x) \) is increasing and the intervals on which \( f(x) \) is decreasing. Then sketch the graph. Add horizontal tangent lines \( f(x)=2 x^{4}-100 x^{2} \) Compute the derivative of \( f(x) \). \( f^{\prime}(x)=\square \)

To find the intervals where \( f(x) = 2x^4 - 100x^2 \) is increasing or decreasing, we first compute its derivative: \[ f'(x) = \frac{d}{dx}(2x^4 - 100x^2) = 8x^3 - 200x. \] Next, we set the derivative equal to zero to find critical points: \[ 8x^3 - 200x = 0 \] Factoring gives: \[ 8x(x^2 - 25) = 0, \] which leads to critical points at: \[ x = 0, \quad x = 5, \quad x = -5. \] Now we analyze the sign of \( f'(x) \) in the intervals determined by these critical points: 1. For \( x < -5 \): Choose \( x = -6 \) → \( f'(-6) = 8(-6)^3 - 200(-6) > 0\) (increasing) 2. For \( -5 < x < 0 \): Choose \( x = -1 \) → \( f'(-1) = 8(-1)^3 - 200(-1) < 0\) (decreasing) 3. For \( 0 < x < 5 \): Choose \( x = 1 \) → \( f'(1) = 8(1)^3 - 200(1) < 0\) (decreasing) 4. For \( x > 5 \): Choose \( x = 6 \) → \( f'(6) = 8(6)^3 - 200(6) > 0\) (increasing) Thus, \( f(x) \) is increasing on the intervals \( (-\infty, -5) \) and \( (5, \infty) \), and decreasing on \( (-5, 0) \) and \( (0, 5) \). Now for sketching the graph of \( f(x) \): - The points \( x = -5 \), \( x = 0 \), and \( x = 5 \) will feature horizontal tangent lines, as these are critical points where \( f'(x) = 0 \). - At \( x = -5 \) and \( x = 5 \), we check if they are local minima or maxima by evaluating the second derivative or using a first derivative test. Overall, the graph will show a steep rise to the left of \( -5 \), a decrease until \( 0 \), and then another decrease until \( 5 \), followed by an increase to the right of \( 5 \). The resulting graph will have its local maximum at \( x = -5 \) and \( x = 5 \) and a local minimum at \( x = 0 \).