Question

Help please!!! (:
f(x)= x^3 / x^2-16 is on the interval [-15, 15]

1.) The function f(x) has a vertical asymptote at x=_____ and x=_____.

2.) f(x) is concave up on the region ____to____ and ____to____.

3.) The inflection point is _____.

Answer 1

Vertical asymptotes exist when the denominator of a fraction is undefined for certain values of x, provided those factors of x do not cancel with the numerator. If a factored term cancels with something in the numerator, then the undefined value in the denominator is only a hole, not an asymptote. Either way, we do not have that worry for this problem. So The places where we are undefined in the denominator are at x = 4 and -4, meaning these are where our asymptotes are. Next we need concacvity and inflection points, which means 2nd derivative. Taking the first derivative requires the quotient rule and will give us [(3x^2)(x^2-16) - (x^3)(2x)]/(x^2-16)^2. SImplifying further I have (x^4 - 48x^2)/(x^2-16)^2. So now I need the quotient rule again for the 2nd derivative. So that gives me [(4x^3 - 96x)(x^2-16)^2 - [(x^4-48x^2)(2)(x^2-16)(2x)]/(x^2-16)^4
Further simplification gives me [32x(x^2+48)]/(x^2-16)^3. So using our numerator, we can see the only inflection point I find by setting the 2nd derivative to 0 is at x = 0. Now to get my concavity, I'll test a point on each side of the inflection point. When trying x = -1, I get a negative numerator and a negative denominator, meaning the function is positive and therefore increasing on the interval -15 to 0. Testing x = 1, I get a positive numerator and a negative denominator, meaning my function is negative and that I am decreasing on the interval from 0 to 15. The steps are long, so I did skip some simplification, but feel free to ask if you want to see the whole process :3

Answer 2

Wow you're explanations were very understanding!(: But for the concave up regions the x values of (-15,0) and (0,15) were wrong. @psymon

Answer 3

With how much typing and sentences are going on, I'm not shocked if I somehow confused myself, lol. I'll relook at it :3

Answer 4

I really appreciate the explanation for every step though, it was easy to understand and follow!(:

Answer 5

Well, in order to find concavity, you would be testing intervals on either side of the inflection points. If only the upward concavity is needed, then I only see that [-15,0] would be concave upward. Is the answer different than that?

Answer 6

it's telling me -15 to 0 is half correct lol -15 is wrong. and same for 0 to 15, 0 is wrong. I'm confused. Could there be another inflection point, or critical numbers?

Answer 7

Half correct? Interesting, lol. Well, I suppose it would be because of the asymptotes in the equation then. Alrighty then, we'll deal with the asymptotes and adjust then. So if we test points on opposite points of asymptotes, too, I can test 4 points. Our asymptotes are the same as in the original function, so at x = 4 and x = -4. So I will first test x = -5. When I do this I have a negative numerator and a positive denominator, meaning the function is negative and therefore decreasing from -15 to -4. Now skipping over the asymptote, I can test x = -1. When I plug that in, I have a negative numerator and a negative denominator, so that would mean from the asymptote of -4 to 0 we are increasing in concavity. Now testing a point on the other side of 0. I can plug in x = 1. Doing this gives me a positive numerator and a negative denominator, makign the function negative and the concavity downward. Finally, I can test a point on the other side of the second asymptote at x = 4. So Ill test the point x = 5. Doing this gives me a positive numerator and denominator, so that makes my function positive and means Im concave upward from 4 to 15. So recap:
(-15, -4) was decreasing
(-4, 0) was increasing
(0, 4) was decreasing and
(4, 15) was increasing.
Basically, it looks like we needed to take into account the asymptotes and not just the inflection points. Now hopefully it'll agree with that answer, lol.

Answer 8

Yup! it worked! and you're explanation help me understand! Thank you so much!(: @psymon

Answer 9

Awesome! A lot of times its the simplifying that gets to be the hard part. Taking a first derivative can often be pretty okay, but the need to simplify in order to get a reasonable 2nd derivaitve is where it gets annoying. Glad ya understood, though, thats good :3

Answer 10

Could you by chance help me with another problem. I understand the process, but I'm getting the wrong answers... @Psymon