Find the intervals on which the graph of \( f \) is concave upward, the intervals on which the graph of \( f \) is concave downward, and the inflection points. \( f(x)=x^{16}+3 x^{2} \)

To determine the concavity of the function \( f(x) = x^{16} + 3x^{2} \), we first need to find the second derivative \( f''(x) \). 1. Calculate the first derivative: \[ f'(x) = 16x^{15} + 6x \] 2. Calculate the second derivative: \[ f''(x) = 240x^{14} + 6 \] To find the intervals of concavity and the inflection points, we need to analyze \( f''(x) \): 3. Set \( f''(x) = 0 \) to find possible inflection points: \[ 240x^{14} + 6 = 0 \\ 240x^{14} = -6 \\ x^{14} = -\frac{1}{40} \] Since \( x^{14} \) is always non-negative, there are no solutions for this equation. Thus, \( f''(x) \) does not change sign. 4. Now, let's analyze the sign of \( f''(x) \): - For all \( x \) (since \( 240x^{14} \) is always non-negative and \( +6 \) is positive), we conclude: - \( f''(x) > 0 \) for all \( x \). From above analysis, we can characteristically conclude: - The graph of \( f \) is concave upward on the entire real line \( (-\infty, \infty) \). - Since there are no points where \( f''(x) = 0 \) or changes sign, there are no inflection points. So, in summary: - The function is concave upward on \( (-\infty, \infty) \). - The function is never concave downward. - There are no inflection points.