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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
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Harold and Kumar?
Just kumar
Harold is in white castle right now

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Other answers:

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

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