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Let f(x)=(x+10)·ln(x+3) for x>-3 . Then f(x) is concave upward on the interval (a,∞) where a =
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For f(x) to be concave up in an interval, f''(x) > 0 in that interval.
I still can't find it. I found the derivative but I am still not seeing the answer.
Hmm what'd you get for your second derivative? This is what I came up with:\[\Large f''(x)\quad=\quad \frac{x-4}{(x+3)^2}\]
Got it! Thank you!
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