Given f'(x) = 1/(3+7x).(adsbygoogle = window.adsbygoogle || []).push({});

Find the interval where the graph of f(x) concaves downward.

First, I differentiate f'(x)= 1/(3+7x) to get f''(x) = -7/(3+7x)^2. So, for the graph to be concave downward, I put f''(x) = -7/(3+7x)^2<0, so 7/(3+7x)^2>0

(3+7x)^2 is always more or equal 0. Then, I exclude -3/7. Therefore, my final answer is that the graph of f(x) concaves downward on interval (-infinity, -3/7) union (-3/7, infinity).

But then again, if we integrate f'(x) = 1/(3+7x) to obtain f(x), we would obtain f(x) = [In (3+7x)]/7 + c.

Therefore, the interval now would be (-3/7, infinity), since the domain of the graph f(x) = [In (3+7x)]/7 is (-3/7, infinity).

But then again, we can integrate f'(x) = 1/(3+7x) to get f(x) = [In (3+7x)]/7, but this time, the term 3+7x is in the modulus form. By this new equation, we can conclude that the graph concave downward on (-infinity, -3/7) union (-3/7, infinity).

Hm... Which solution is the correct one ?

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# Homework Help: Confusion of Concavity

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