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anonymous
 4 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=___
anonymous
 4 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=___

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Harold is in white castle right now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can Kumar be of assistance?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Depends on the situation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The situation above ^^^ lols

across
 4 years ago
Best 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 \(x>10\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what rules did you use?

across
 4 years ago
Best ResponseYou've already chosen the best response.3I just differentiated the function twice.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, I know. Could you be more specific?

across
 4 years ago
Best 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.

across
 4 years ago
Best ResponseYou've already chosen the best response.3That's the product rule.

across
 4 years ago
Best ResponseYou've already chosen the best response.3With the chain rule applied to the parentheses.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok. thanks. you're really pretty by the way. :)
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