Find tie inteivals on which \( f(x) \) is increasing and the in \( f(x)=2 x^{4}-100 x^{2} \) Compute the derivative of \( f(x) \). \( f^{\prime}(x)=\square \)

To determine the intervals where \( f(x) \) is increasing, we first compute the derivative of the function \( f(x) = 2x^4 - 100x^2 \). The derivative is calculated as follows: \[ f'(x) = \frac{d}{dx}(2x^4 - 100x^2) = 8x^3 - 200x \] Now, to find the intervals where \( f(x) \) is increasing, we need to set the derivative greater than zero: \[ 8x^3 - 200x > 0 \] Factoring out what we can gives us: \[ 8x(x^2 - 25) > 0 \implies 8x(x - 5)(x + 5) > 0 \] The critical points are \( x = -5, 0, 5 \). We can use these points to test the sign of the derivative in the intervals created: 1. **Interval \( (-\infty, -5) \)**: Choose \( x = -6 \) → \( f'(-6) = 8(-6)(-6^2 - 25) > 0 \) (Positive). 2. **Interval \( (-5, 0) \)**: Choose \( x = -1 \) → \( f'(-1) = 8(-1)(-1^2 - 25) < 0 \) (Negative). 3. **Interval \( (0, 5) \)**: Choose \( x = 1 \) → \( f'(1) = 8(1)(1 - 25) < 0 \) (Negative). 4. **Interval \( (5, \infty) \)**: Choose \( x = 6 \) → \( f'(6) = 8(6)(36 - 25) > 0 \) (Positive). Thus, \( f(x) \) is increasing on the intervals \( (-\infty, -5) \) and \( (5, \infty) \). In conclusion: - The derivative is \( f'(x) = 8x^3 - 200x \). - \( f(x) \) is increasing on \( (-\infty, -5) \) and \( (5, \infty) \).