Question

I would really appreciate help finding the concavity for this function, i can't seem to get it.
f(x) = { x^3 }/{ x^2 - 25 }

Answer 1

i can see why because the second derivative sucks to find. but once you have it, it is easy.

Answer 2

ok i lied, because i did it wrong. it is still a pain

Answer 3

\[f''(x)=\frac{50x(x^2+75)}{(x^2-25)^3}\]

Answer 4

\[x^2+75>0\] so we can ignore that part
\[(x^2-25)^3> 0\iff x^2-25>0\] so we can forget about the cube. we only need to consider this
\[\frac{x}{x^2-25}=\frac{x}{(x+5)(x-5)}\] which will change sign at
\[-5,0,5\] so we need to consider the intervals
\[(-\infty, -5),(-5,0),(0,5), (5,\infty)\]

Answer 5

it will alternate sign over these intervals, and you can check that it is positive on
\[(-5,0)\cup (5,\infty)\] and negative on the other two. so those are the intervals over which your function is concave up and it will be concave down on
\[(-\infty,-5)\cup (0,5)\]

Answer 6

I have to mention too that its defined on the interval (-20,18)

Answer 7

does that make any difference?

Answer 8

then adjust accordingly which just means change minus infinity to -20 and infinity to 18

Answer 9

so the concavity would be up on -5,0 and also on 5,18 and then it would be down on -20,-5 and 0,5?

Answer 10

I got it thank you so much!