Find the points of inflection and discuss the concavity of the function $f ( x ) = x \sqrt { x + 16 }$
A) no inflection points; concave up on $( - 16 , \infty )$
B) no inflection points; concave down on $( - 16 , \infty )$
C) inflection point at $x = 16$; concave up on $( - 16 , \infty )$
D) inflection point at $x = 0$; concave up on $( - 16,0 )$; concave down on $( 0 , \infty )$
E) inflection point at $x = 16$; concave down on $( - 16 , \infty )$

Question

Quizplus · Accepted Answer

To determine the points of inflection and concavity, we need to find the second derivative of $f(x) = x\sqrt{x + 16}$ and analyze its sign. First, let's find the first derivative $f'(x)$ using the product rule and chain rule:$$f(x) = x(x + 16)^{1/2}$$$$f'(x) = (x + 16)^{1/2} + \frac{1}{2}x(x + 16)^{-1/2}$$$$f'(x) = \sqrt{x + 16} + \frac{x}{2\sqrt{x + 16}}$$Now, let's find the second derivative $f''(x)$:$$f''(x) = \frac{1}{2\sqrt{x + 16}} - \frac{x}{4(x + 16)^{3/2}} + \frac{1}{2\sqrt{x + 16}}$$$$f''(x) = \frac{1}{\sqrt{x + 16}} - \frac{x}{4(x + 16)^{3/2}}$$$$f''(x) = \frac{4(x + 16) - x}{4(x + 16)^{3/2}}$$$$f''(x) = \frac{3x + 64}{4(x + 16)^{3/2}}$$Since $x + 16 > 0$ for $x > -16$, the denominator is always positive. The numerator $3x + 64$ is also always positive for $x > -16$. Therefore, $f''(x) > 0$ for all $x > -16$, indicating the function is concave up on $(-16, \infty)$ and there are no inflection points because the sign of $f''(x)$ does not change.

Find the Points of Inflection and Discuss the Concavity of the Function