Question

Help Please!
f(x)=5x^5-7x^4

1.) Use interval notation to indicate where f(x) is decreasing.
2.) Use interval notation to indicate where f(x) is concave up and concave down.
3.) Find all x values of the inflection points of f. If there is any.
4.) Find all horizontal and vertical asymptotes of f. If there is any.

----I understand the process, but my numbers are coming out wrong..

Answer 1

Yeah, I'll look it over :3

Answer 2

Thank you!!! (:

Answer 3

So immediately looking at #4, we can throw out any asymptotes. Asymptotes would require us to have a denominator, which we do not have one. Increasing and decreasing indicates the first derivative then set = to 0. So the first derivative would be 25x^4 - 28x^3. Now I need to factor and set the derivative to 0. All I can really do with this function, though, is factor out an x^3, which would leave me with x^3(25x-28). This will leave me with 2 critical points. Seeting each of these factors to 0 gives me x^3 = 0 and 25x - 28 = 0. Solving for x then gives me the critical points of x = 0 and x = 28/25. Now if I want to know where the function is increasing or decreasing, I choose points on each side of our 2 critical points. to the left of x = 0, I can test x= -1, making sure I plug it back into the first derivative and not the original equation. So if I plug in -1 into the derivative, I get a positive answer, meaning from (-∞, 0) I'm increasing. Now I can test x = 1 and see what that gives me. Setting x equal to 1, the derivative is negative, meaning from (0, 28/25) my function is decreasing. Now I can test the last point, which I wil pick 2 for. if I use x = 2, I get a positive answer, meaning from (28/25, ∞) I'm increasing. Note that I only use parenthesis for these answers. I do not use brackets because these points are were the tangent lines of the graph are 0, basically meaning there cannot be any sort of increasing or decreasing at these points. So the question asks for only increasing, so I would put (-∞, 0) and (28/25, ∞). So now I need inflection points and concavity, meaning now I need the 2nd derivative. Taking the 2nd derivative, I get 100x^3 - 84x^2. Setting this equal to 0 will give me inflection points. So in order to factor this, I will factor out a 4x^2, leaving me with 4x^2(25x - 21). Setting each factor = to 0 now, I have 4x^2 = 0 and 25x - 21 = 0. So this gives me 2 inflection points of x = 0 and x = 21/25. Now just like I did for increasing and decreasing, I do the same type of interval testing for concavity. The first point I will test is x = -1. If I plug in -1, I get a negative answer, meaning I am concave downward from (-∞, 0). Now I will test (1/2) for my next test point (since I can't use a nicer number :( . When I try (1/2), I get a negative answer again, meaning i"m concave downward on this interval as well. So finally I will try the last interval using x = 1. When I use x = 1, I get a positive answer meaning I am concave upward on this interval.
So in all, I am concave downward on (-∞, 0) and (0, 21/25) and concave upward on (21/25, ∞)
I had inflection points at 0 and 21/25
I had critical points at 0 and 28/25
And finally, I was decreasing from (0, 28/25)

Answer 4

I got the same inflection points and they were wrong... :( @Psymon

Answer 5

And I found where I went wrong.. haha I factored wrong which made it all ugly..

Answer 6

Lol, I see. Yeah, I was trying to see what was up with the inflection points not being correct. But yeah, the factoring should come out pretty clean. Factoring the 2nd derivative leaves you with 4x^2(25x - 21). That being said, not sure what other values could be inflection points if not 0 and 21/25.

Answer 7

I figured it out. I guess 0 isn't an inflection point but 21/25 is. Thanks for all the help once again! I really appreciate it! (: @Psymon

Answer 8

Ah, okay. Yeah, I just wanted to verify the reasoning. When you have the same concavity behavior on both sides of an inflection point, it actually means said point is not a true inflection point :3