Interval increasing/decreasing

[jwxie]

Homework Statement

ln(3x+5) Determine intervals on which the function is increasing, decreasing, concave up, and concave down.

Homework Equations

The Attempt at a Solution

So I did f '(x) = 3/3x+5
this gives me 3x+5 = 0, and I get x = -5/3 (point where the y is zero)

Now, I did f '' (x), and got -3/(3x+5)^2
this gives me (3x+5)(3x+5), i did a number line test, with the only value -5/3, and anything below -5/3 is negative, and before -5/3 is positive

this gives me that, the function is decreasing after -5/3, and increasing after -5/3.

Now what about the concavity?

 

[Mark44]

Homework Statement

ln(3x+5) Determine intervals on which the function is increasing, decreasing, concave up, and concave down.

Homework Equations

The Attempt at a Solution

So I did f '(x) = 3/3x+5
this gives me 3x+5 = 0, and I get x = -5/3 (point where the y is zero)

No, f'(x) is never zero.

Now, I did f '' (x), and got -3/(3x+5)^2
this gives me (3x+5)(3x+5), i did a number line test, with the only value -5/3, and anything below -5/3 is negative, and before -5/3 is positive

this gives me that, the function is decreasing after -5/3, and increasing after -5/3.

You're not taking into account the domain of the original function, f(x) = ln(3x + 5). This function is defined only for x such that 3x + 5 > 0. The first and second derivatives have the same domain.

Now what about the concavity?

 

[jwxie]

I meant -5/3 is where the vertical asymto exists.
http://www.wolframalpha.com/input/?i=+ln(3x+5)

so how would you describe its concavities?
I had the right picture.

 

[Mark44]

Look at the sign of f''(x), which by the way is not equal to 3/(3x + 5)^2. Keep in mind what the domain is.