anonymous
  • anonymous
where is f concave up if f"(x)=(2x)(x^2+27)/(x^2-9)^3?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
where ever \[ (2x)(x^2+27)/(x^2-9)^3>0\]
anonymous
  • anonymous
not as hard as it seems, because \[x^2+27>0\] always, so we can ignore that factor
anonymous
  • anonymous
\[2x>0\iff x>0\]

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anonymous
  • anonymous
\[(x^2-9)^3>0\iff x^2-9>0\iff x<-3 \text{ or } x>3\]
anonymous
  • anonymous
putting it all together we get that \[f''(x)>0\] if \[-33\]
anonymous
  • anonymous
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
thanks, it makes a lot more sense now. But what happened to the 3 in\[(x ^{2}-9)^3\]? how did we get to \[x^2-9>0\]

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