anonymous
  • anonymous
Let f(x)=(x+12) * ln(x+1) for x>-1. Then f(x)is concave upward on the interval (a,infinity) where a=___
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Harold and Kumar?
anonymous
  • anonymous
Just kumar
anonymous
  • anonymous
Harold is in white castle right now

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anonymous
  • anonymous
Can Kumar be of assistance?
anonymous
  • anonymous
Depends on the situation.
anonymous
  • anonymous
The situation above ^^^ lols
across
  • across
\[f(x)=\ln(x+1)(x+12)\]\[f'(x)=\frac{x+12}{x+1}+\ln(x+1)\]\[f''(x)=\frac{x-10}{(x+1)^2}.\]It follows that \(f''(x)>0\) when \(x>10\).
anonymous
  • anonymous
what rules did you use?
across
  • across
I just differentiated the function twice.
anonymous
  • anonymous
Yeah, I know. Could you be more specific?
across
  • across
\[\frac{d}{dx}\left[\ln(x+1)(x+12)\right]=\frac{d}{dx}\left[\ln(x+1)\right](x+12)+\ln(x+1)\frac{d}{dx}\left[(x+12)\right]\]\[=\frac{x+12}{x+1}\frac{d}{dx}\left[x+1\right]+\ln(x+1)\cdot1=\frac{x+12}{x+1}\cdot1+\ln(x+1)\]\[\frac{x+12}{x+1}+\ln(x+1)\]Then do that a second time.
across
  • across
That's the product rule.
across
  • across
With the chain rule applied to the parentheses.
anonymous
  • anonymous
ok. thanks. you're really pretty by the way. :)

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