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mariomintchev Group Title

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
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  1. mariomintchev Group Title
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    Harold and Kumar?

    • 2 years ago
  2. kumar2006 Group Title
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    Just kumar

    • 2 years ago
  3. kumar2006 Group Title
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    Harold is in white castle right now

    • 2 years ago
  4. mariomintchev Group Title
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    Can Kumar be of assistance?

    • 2 years ago
  5. kumar2006 Group Title
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    Depends on the situation.

    • 2 years ago
  6. mariomintchev Group Title
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    The situation above ^^^ lols

    • 2 years ago
  7. across Group Title
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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 \(x>10\).

    • 2 years ago
  8. mariomintchev Group Title
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    what rules did you use?

    • 2 years ago
  9. across Group Title
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    I just differentiated the function twice.

    • 2 years ago
  10. mariomintchev Group Title
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    Yeah, I know. Could you be more specific?

    • 2 years ago
  11. across Group Title
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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.

    • 2 years ago
  12. across Group Title
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    That's the product rule.

    • 2 years ago
  13. across Group Title
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    With the chain rule applied to the parentheses.

    • 2 years ago
  14. mariomintchev Group Title
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    ok. thanks. you're really pretty by the way. :)

    • 2 years ago
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