Really annoying math issues

[Xenus]

Let f(x)=(9x)/(x^2+9)


I got the points of inflection are -3,0,3 and that it's Decreasing and concave up on [3,I) and Decreasing and concave down on (-I,-3] but it tells me this in wrong but the others [-3,0], [0,3] it said is right so these would seem to have to be right.

 

[Basic]

I assume that you meant: f(x) = 9*x/(x^2+9)

A point of inflection is where the function switch between convex and concave. Ie, where the second derivative cross zero. That doesn't happen at -3 and 3. Maybe you made a mistake in the derivative.

Btw in (A) you ask for an increasing interval, and then you write about a decreasing interval. Was that intensional?

 

[Xenus]

Yeah that's what I mean't what I was asking for when the function is concave up and decreasing at the sam time and for when it is decreaind and concave down at the same time

 

[suryad]

TI-89.....nuff said.....

If you want to confirm your answer that is....or Maple....or Mathematica....or Matlab.

 

[Basic]

Decreasing => f'(x) < 0
Concave up => f''(x) > 0

Find stationary points: f'(x) = 0 => x = -3, 3
Find inflection points: f''(x) = 0 => x = 0, two other values not equal to -3, 3

You've either mixed up f'(x) and f''(x), or calculated f''(x) wrong.
Show your f'(x) and f''(x).

 

[Xenus]

nah I was doing it the lazy way and throwing the eqaution into the calculator and graphing it

 

[AlNom]

As x goes to infinity, the function behaves as a 1/x function.

i.e. x^2 >> 9

f(x)~ 9x/x^2 ~ 1/x

So probably at some point you find your missing inflection points and then the curve goes concave up. The same goes for x going to negative infinity where it behaves as -1/x.


edit: to further see the approximation, let's take x=100 billion

f= 9*10^11/(10^22+9)

and then compare it to:

9/10^11


...and your calculator probably won't even bother making a distinction.