Question

Find the open intervals where the function

f(x)= ln (x2+36) is concave upward or concave downward. Find any inflection points.
1)Select the correct choice below and fill in any answer boxes within your choice.​
A) The function has a point of inflection at:
B) The fuction does not have an inflection point
2) Select the correct choice below and fill in any answer boxes within your choice.
A) The function is concave upward on the​ interval(s):
B) The function is never concave upward

Answer 1

The function f(x) = ln(x^2 + 36) is concave upward for all real numbers except at x = ± 6. However, there are no points of inflection because the concavity does not change at those points.

Step-by-step explanation:

To determine the concavity of the function f(x) = ln(x2 + 36), we need to find its second derivative and analyze the sign.

First, find the first derivative f'(x):

1. f'(x) = \(\frac{d}{dx}[ln(x2 + 36)] = \frac{2x}{x2 + 36}

Next, find the second derivative f''(x):

1. f''(x) = \(\frac{d}{dx}[\frac{2x}{x2 + 36}] = \frac{2(x2 + 36) - 2x(2x)}{(x2 + 36)2} = \frac{72 - 2x2}{(x2 + 36)2}

To determine intervals of concavity, we look at the sign of f''(x). The function is concave upward where f''(x) > 0 and concave downward where f''(x) < 0. The possible points of inflection are where f''(x) changes sign, which occurs when the numerator equals zero. The numerator 72 - 2x2 is zero when x2 = 36, or x = ± 6. Test the intervals to determine the concavity:

• Test interval (-∞, -6): Choose x = -7, f''(-7) > 0, so concave upward.
• Test interval (-6, 6): Choose x = 0, f''(0) > 0, so concave upward.
• Test interval (6, ∞): Choose x = 7, f''(7) > 0, so concave upward.

Thus, f(x) is concave upward for all x ≠ -6 and 6, and there are no points of inflection, because the concavity does not change at x = ± 6.