## mariomintchev 3 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=___

1. mariomintchev

Harold and Kumar?

2. kumar2006

Just kumar

3. kumar2006

Harold is in white castle right now

4. mariomintchev

Can Kumar be of assistance?

5. kumar2006

Depends on the situation.

6. mariomintchev

The situation above ^^^ lols

7. across

$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 $$x>10$$.

8. mariomintchev

what rules did you use?

9. across

I just differentiated the function twice.

10. mariomintchev

Yeah, I know. Could you be more specific?

11. across

$\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.

12. across

That's the product rule.

13. across

With the chain rule applied to the parentheses.

14. mariomintchev

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