Answer to Question #117986 in Calculus for sam

Question

Answer the following questions for the function

f(x)=x3x2−36

defined on the interval [−15,15].

a.) Enter the x-coordinates of the vertical asymptotes of f(x) as a comma-separated list. That is, if there is just one value, give it; if there are more than one, enter them separated commas; and if there are none, enter NONE .

Answer:

-6,6

b.) f(x) is concave up on the region

(-6,0)U(6,18)

Note: Give your answer in interval notation.

c.) Enter the x-coordinates of the inflection point(s) for this function as a comma-separated list.

Answer:

Answer 1

a) Function $f(x) = \dfrac{x^3}{x^2-6} = \dfrac{x^3}{(x-6)(x+6)}.$ Vertical asymptotes correspond to points, in which $x^2-6=0$ , so the answer is

-6,6

b) to determine the interval of concavity we should determine the interval of the positive sigh of the second derivative.

$f'(x) = \dfrac{x^4-108 x^2}{x^4-72 x^2+1296}$ , $f''(x) = \dfrac{72 x^3+7776 x}{x^6-108x^4+3888x^2-46656} = \dfrac{x(72x^2+7776)}{(x+6)^3(x-6)^3}$ .

The last function if positive for $x\in(-6,0)\cup(6;15).$

c) Inflection point (see https://en.wikipedia.org/wiki/Inflection_point) is a point on continuous plane curve at which the curve changes her concavity. This curve is continuous on the intervals (-15,-6), (-6,6), (6,15). $f''(x) = 0$ only at $x=0$ , so it is the inflection point.

We can also look at the graph of this function and see that it has only one inflection point on (-15,15)