Questions: 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 + 7x^2 For what interval(s) of x is the graph of f concave upward? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. (-∞, ∞) (Type your answer in interval notation. Type an exact answer. Use a comma to separate answers as needed.) B. The graph is never concave upward. For what interval(s) of x is the graph of f concave downward? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. (Type your answer in interval notation. Type an exact answer. Use a comma to separate answers as needed.) B. The graph is never concave downward.

Solution Steps

To solve the problem, we need to determine the concavity of the function \( f(x) = x^{16} + 7x^2 \) by finding the second derivative and analyzing its sign.

The first derivative of the function \( f(x) \) is found by differentiating each term:

\[ f'(x) = \frac{d}{dx}(x^{16}) + \frac{d}{dx}(7x^2) = 16x^{15} + 14x \]

Next, we find the second derivative by differentiating \( f'(x) \):

\[ f''(x) = \frac{d}{dx}(16x^{15}) + \frac{d}{dx}(14x) = 240x^{14} + 14 \]

The concavity of the function is determined by the sign of \( f''(x) \).

• Concave Upward: The graph is concave upward where \( f''(x) > 0 \).
• Concave Downward: The graph is concave downward where \( f''(x) < 0 \).

The expression for the second derivative is:

\[ f''(x) = 240x^{14} + 14 \]

Since \( x^{14} \) is always non-negative for all real \( x \) (as even powers of any real number are non-negative), and \( 240x^{14} \) is non-negative, the entire expression \( 240x^{14} + 14 \) is always positive. Therefore, \( f''(x) > 0 \) for all \( x \).

• Concave Upward: Since \( f''(x) > 0 \) for all \( x \), the graph is concave upward on the interval \((-\infty, \infty)\).
• Concave Downward: There are no intervals where \( f''(x) < 0 \), so the graph is never concave downward.

Inflection points occur where \( f''(x) = 0 \) or where \( f''(x) \) changes sign. Since \( f''(x) = 240x^{14} + 14 \) is always positive and never zero, there are no inflection points.

Final Answer