mariomintchev
Let f(x)=(x+12) * ln(x+1)
for x>-1. Then f(x)is concave upward on the interval (a,infinity) where a=___
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mariomintchev
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Harold and Kumar?
kumar2006
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Just kumar
kumar2006
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Harold is in white castle right now
mariomintchev
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Can Kumar be of assistance?
kumar2006
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Depends on the situation.
mariomintchev
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The situation above ^^^
lols
across
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\[f(x)=\ln(x+1)(x+12)\]\[f'(x)=\frac{x+12}{x+1}+\ln(x+1)\]\[f''(x)=\frac{x-10}{(x+1)^2}.\]It follows that \(f''(x)>0\) when \(x>10\).
mariomintchev
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what rules did you use?
across
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I just differentiated the function twice.
mariomintchev
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Yeah, I know. Could you be more specific?
across
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\[\frac{d}{dx}\left[\ln(x+1)(x+12)\right]=\frac{d}{dx}\left[\ln(x+1)\right](x+12)+\ln(x+1)\frac{d}{dx}\left[(x+12)\right]\]\[=\frac{x+12}{x+1}\frac{d}{dx}\left[x+1\right]+\ln(x+1)\cdot1=\frac{x+12}{x+1}\cdot1+\ln(x+1)\]\[\frac{x+12}{x+1}+\ln(x+1)\]Then do that a second time.
across
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That's the product rule.
across
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With the chain rule applied to the parentheses.
mariomintchev
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ok. thanks. you're really pretty by the way. :)