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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=___
 2 years ago
 2 years ago
Let f(x)=(x+12) * ln(x+1) for x>1. Then f(x)is concave upward on the interval (a,infinity) where a=___
 2 years ago
 2 years ago

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mariomintchevBest ResponseYou've already chosen the best response.1
Harold and Kumar?
 2 years ago

kumar2006Best ResponseYou've already chosen the best response.0
Harold is in white castle right now
 2 years ago

mariomintchevBest ResponseYou've already chosen the best response.1
Can Kumar be of assistance?
 2 years ago

kumar2006Best ResponseYou've already chosen the best response.0
Depends on the situation.
 2 years ago

mariomintchevBest ResponseYou've already chosen the best response.1
The situation above ^^^ lols
 2 years ago

acrossBest ResponseYou've already chosen the best response.3
\[f(x)=\ln(x+1)(x+12)\]\[f'(x)=\frac{x+12}{x+1}+\ln(x+1)\]\[f''(x)=\frac{x10}{(x+1)^2}.\]It follows that \(f''(x)>0\) when \(x>10\).
 2 years ago

mariomintchevBest ResponseYou've already chosen the best response.1
what rules did you use?
 2 years ago

acrossBest ResponseYou've already chosen the best response.3
I just differentiated the function twice.
 2 years ago

mariomintchevBest ResponseYou've already chosen the best response.1
Yeah, I know. Could you be more specific?
 2 years ago

acrossBest ResponseYou've already chosen the best response.3
\[\frac{d}{dx}\left[\ln(x+1)(x+12)\right]=\frac{d}{dx}\left[\ln(x+1)\right](x+12)+\ln(x+1)\frac{d}{dx}\left[(x+12)\right]\]\[=\frac{x+12}{x+1}\frac{d}{dx}\left[x+1\right]+\ln(x+1)\cdot1=\frac{x+12}{x+1}\cdot1+\ln(x+1)\]\[\frac{x+12}{x+1}+\ln(x+1)\]Then do that a second time.
 2 years ago

acrossBest ResponseYou've already chosen the best response.3
That's the product rule.
 2 years ago

acrossBest ResponseYou've already chosen the best response.3
With the chain rule applied to the parentheses.
 2 years ago

mariomintchevBest ResponseYou've already chosen the best response.1
ok. thanks. you're really pretty by the way. :)
 2 years ago
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