When Does the Function f(x) Concave Downward?

[lkh1986]

Given f'(x) = 1/(3+7x).
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 ?

 

[radou]

...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 ?

Remember that f(x) = lnx is defined for x > 0.

 

[Office_Shredder]

\int \frac{1}{x}dx = \ln(|x|)

so it would be defined for x less than zero

EDIT: My absolute value bars don't seem to be coming through... :(

 

[lkh1986]

\int \frac{1}{x}dx = \ln(|x|)

so it would be defined for x less than zero. (

Oh, yeah, then if I apply this formula I'll get the answer for the interval is (-infinity,-3/7) union (-3/7, infinity), right?

 

[Office_Shredder]

That looks right to me.

Damnit! Now the absolute value bars are showing up... I swear, I'm going crazy or something