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

Question

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 · Answer

Harold and Kumar?

anonymous · Answer

Just kumar

anonymous · Answer

Harold is in white castle right now

anonymous · Answer

Can Kumar be of assistance?

anonymous · Answer

Depends on the situation.

anonymous · Answer

The situation above ^^^
lols

across · Answer

$$f(x)=\ln(x+1)(x+12)$$$$f'(x)=\frac{x+12}{x+1}+\ln(x+1)$$$$f''(x)=\frac{x-10}{(x+1)^2}.$$It follows that $f''(x)>0$ when $x>10$.

anonymous · Answer

what rules did you use?

across · Answer

I just differentiated the function twice.

anonymous · Answer

Yeah, I know. Could you be more specific?

across · Answer

$$\frac{d}{dx}\left[\ln(x+1)(x+12)\right]=\frac{d}{dx}\left[\ln(x+1)\right](x+12)+\ln(x+1)\frac{d}{dx}\left[(x+12)\right]$$$$=\frac{x+12}{x+1}\frac{d}{dx}\left[x+1\right]+\ln(x+1)\cdot1=\frac{x+12}{x+1}\cdot1+\ln(x+1)$$$$\frac{x+12}{x+1}+\ln(x+1)$$Then do that a second time.

across · Answer

That's the product rule.

across · Answer

With the chain rule applied to the parentheses.

anonymous · Answer

ok. thanks. you're really pretty by the way. :)