Question

Answer the following questions for the function f(x)= (x^3)/(x^2-9) defined on the interval [-17,15]. Enter points, such as inflection points in ascending order, i.e. smallest x values first.
A. The function f(x) has vertical asymptotes? (2 of them)
B. f(x) is concave up on the region () to () and () to ().
C. The inflection point for this function is?

Answer 1

Answer on Question #43469-Engineering-Other

Answer the following questions for the function $f(x) = \frac{(x^3)}{x^2 - 9}$ defined on the interval [-17,15]. Enter points, such as inflection points in ascending order, i.e. smallest x values first.

A. The function $f(x)$ has vertical asymptotes? (2 of them)

B. $f(x)$ is concave up on the region () to () and () to ().

C. The inflection point for this function is?

Solution

A. The function has vertical asymptotes if it is reduced (no common factors in numerator and denominator) and the denominator is zero.

Thus $x^{2} - 9 \to x = \pm 3$. The vertical asymptotes are at $x = -3$ and $x = 3$.

B. The function is concave up on an interval where the 2nd derivative is positive.

$$f(x) = \frac{(x^3)}{x^2 - 9}.$$ $$f'(x) = \frac{(x^2 - 9) \cdot 3x^2 - x^3 \cdot 2x}{(x^2 - 9)^2} = \frac{x^4 - 27x^2}{(x^2 - 9)^2}.$$ $$\begin{array}{l} f''(x) = \frac{(x^4 - 18x^2 + 81) \cdot (4x^3 - 27 \cdot 2x) - (x^4 - 27x^2)(4x^3 - 36x)}{(x^2 - 9)^4} \\ = \frac{18x^5 + 324x^3 - 4374x}{(x^2 - 9)^4} = \frac{18(x^5 + 18x^3 - 243x)}{(x^2 - 9)^4} = \frac{18x(x^2 - 9)(x^2 + 27)}{(x^2 - 9)^4} \\ = \frac{18x(x^2 + 27)}{(x^2 - 9)^3}. \end{array}$$

$f''(x) > 0$ on $(-3,0)$ and $(3,\infty)$ so the given function is concave up on $(-3,0)$ and $(3,15)$.

C. The inflection point occurs when $f''(x) = 0$ and the sign of the second derivative changes parity. The second derivative is zero at $x = 0$ which is the inflection point.

http://www.AssignmentExpert.com/