Question

where is f concave up if f"(x)=(2x)(x^2+27)/(x^2-9)^3?

Answer 1

where ever
\[ (2x)(x^2+27)/(x^2-9)^3>0\]

Answer 2

not as hard as it seems, because
\[x^2+27>0\] always, so we can ignore that factor

Answer 3

\[2x>0\iff x>0\]

Answer 4

\[(x^2-9)^3>0\iff x^2-9>0\iff x<-3 \text{ or } x>3\]

Answer 5

putting it all together we get that
\[f''(x)>0\] if
\[-3<x<0\] or
\[x>3\]

Answer 6

ok, so basically when i have problems like these first i have to set everything to > if it is asking me for concave up and < if it is asking for concave down, unless i know that one of the quantities is going to be positive..like the (x^2+27). and that one we can ignore because we know it will be positive since it's squared?

Answer 7

yes, your entire job here is to know that the function is concave up if
\[f''(x)>0\] and that is what you have to solve
it is often a pain

Answer 8

yes you can ignore the factor
\[x^2+27\] because
\[x^2\geq 0\] so
\[x^2+27\geq 27\] i.e. that term does not contribute to the sign of the function

Answer 9

thanks, it makes a lot more sense now. But what happened to the 3 in\[(x ^{2}-9)^3\]? how did we get to \[x^2-9>0\]