Find the intervals on which the graph of I is concave upward, the intorvals on which the graph of \( f \) is concave downwarc, and the inflection points \[ f(x)=x^{15}+3 x^{2} \] For what interval(s) of x is the graph of f concavo upward? Seloct the correct choica below and, if necessary, fill in the answer box to complete your choice. A. \( (-\infty, \infty) \) (Type your answer in inlerval notation. Type an exact answor. Uso a comma to soparate answers as needed.) B. The graph is never concavo upward. For what interva(s) of \( x \) is the graph of f concavo downward? Select the conect choice beiow and, if necessary, fill in the answer box to compete your choice. A \( \square \) (Type your answor in inlerval notation Type an exacl answer. Use a comma to soparato answors as noedod.) B. The graph is never concave downward. Determine the \( \times \) coordinates of any inflection points of the graph of \( I(x) \). Select the correct choice below and, if necessary, fil in the answer box to complete your cheico. A. \( \square \) (Type an exacl answer. Use a comma to separate answers as needed.) B. There are no inflection points.

Let \[ f(x)=x^{15}+3x^{2}. \] **Step 1. Compute the second derivative** Differentiate twice. First, the first derivative is \[ f'(x)=15x^{14}+6x. \] Then the second derivative is \[ f''(x)=15 \cdot 14 \, x^{13}+6=210x^{13}+6. \] **Step 2. Set up the concavity inequality** The concavity is determined by the sign of \( f''(x) \): - \(f\) is concave upward when \(f''(x) > 0\). - \(f\) is concave downward when \(f''(x) < 0\). We have \[ 210x^{13}+6>0. \] Subtract 6: \[ 210x^{13}>-6. \] Divide by 210 (since 210 is positive, the inequality sign does not change): \[ x^{13}>-\frac{6}{210}=-\frac{1}{35}. \] Because the function \( x^{13} \) is strictly increasing (since 13 is odd), the inequality \[ x^{13}>-\frac{1}{35} \] holds exactly when \[ x> -\sqrt[13]{\frac{1}{35}}, \] or equivalently, \[ x> -35^{-1/13}. \] **Step 3. Determine the intervals for concavity** - The graph is concave upward when \[ x> -35^{-1/13}, \] so in interval notation: \[ \left(-35^{-1/13},\infty\right). \] - Similarly, \(f\) is concave downward when \[ x^{13}<-\frac{1}{35} \] that is, when \[ x< -35^{-1/13}, \] so in interval notation: \[ \left(-\infty,-35^{-1/13}\right). \] **Step 4. Find the inflection point(s)** An inflection point occurs when \( f''(x)=0 \) and the concavity changes. Set the second derivative equal to zero: \[ 210x^{13}+6=0 \quad \Longrightarrow \quad 210x^{13}=-6 \quad \Longrightarrow \quad x^{13}=-\frac{6}{210}=-\frac{1}{35}. \] Taking the 13th root on both sides (which is valid since 13 is odd), we obtain: \[ x=-\sqrt[13]{\frac{1}{35}} = -35^{-1/13}. \] Since the sign of \( f''(x) \) changes from negative (for \( x<-35^{-1/13} \)) to positive (for \( x>-35^{-1/13} \)), there is an inflection point at \[ x=-35^{-1/13}. \] **Final Answers** - For what interval(s) of \( x \) is the graph of \( f \) concave upward? Answer: \(\displaystyle \left(-35^{-1/13},\infty\right)\). - For what interval(s) of \( x \) is the graph of \( f \) concave downward? Answer: \(\displaystyle \left(-\infty,-35^{-1/13}\right)\). - The \( x \)-coordinate(s) of the inflection point(s) of \( f \) is: Answer: \( x=-35^{-1/13} \).