Question

f(x)=9x-3ln(x), x>0

Use interval notation to indicate where f(x) is increasing and decreasing.
Use interval notation to indicate where f(x) is concave up.

Answer 1

How shall we determine increasing or decreasing? Do we gt to use calculus?

Answer 2

f'(x) = 9 - 3/x
9 - 3/x = 0 --> x = 1/3
x = 1/3 is the critical point, therefore where the function changes from increase to decreasing or vice versa.
Plugging in x = 1/6 into f': f'(1/6) = 9 - 3/(1/6) = 9-18 = -18
Therfore for 0<x<1/3: f(x) is decreasing.

Plug x = 2/3 into f': f'(2/3) = 9 - 3/(2/3) = 9 - 9/2 = 18/2 - 9/2 = 9/2
Thefore for x>(1/3), f(x) is increasing.

Now find f double prime f''(x) = 3/(x^2)
To find where it is concave up we need to find x where f''(x) is greater than or equal to zero. f''(x) cannot go below zero without being negative and because our interval is x>0 this is not possible. Thus, f(x) is concave up for x > 0

Answer 3

Sorry, I don't know if you are familiar with prime notation, it is used to signify the order of the derivative. Thus the derivative of f(x) (f of x) would be f prime of x, which is signified by f'(x). Similarily, for the second derivative the notation would include two apostrophes f''(x).

Answer 4

Yeah, I understood that. I'm trying to understand your notation you used for the increasing/decreasing. I'm not familiar with that.

Answer 5

okay, I got decreasing.. But, I'm still stuck on increasing haha

Answer 6

Ahhh ok. Basically i took the derivative of the function which I noted as f'(x) = 9 - 3/x. This finds me the slope of f(x) at any x. Since we're finding where it might be increasing/decreasing, it is useful to find the point at which slope of f(x) thus f'(x) would be zero (also known as a critical point), thus I set the derivative equal to zero and solved for x.
I'll do it in the equation thing so it is easier to understand:
\[f(x) = 9x - 3*ln(x)\]
The derivative of which is: \[f'(x) = 9 - 3/x\]
Critical point:
\[9 - \frac{ 3 }{ x } = 0\]\[9 = \frac{ 3 }{ x }\]\[x = 1/3\]
Now i'm looking on both sides of the critical point. So I take two random numbers, in my case 1/6 (less than 1/3) and 2/3 (greater than 2/3).
Since we know that below 1/3 it is decreasing, we are now testing if above 1/3 it is increasing. To test this we plug 2/3 into our derivative and we hope to find a positive number. \[f'(2/3) = 9 - \frac{ 3 }{ \frac{ 2 }{ 3 } } = 9 -\frac{ 9 }{ 2 } = \frac{ 18 }{ 2 }-\frac{ 9 }{ 2 }=\frac{ 9 }{ 2 }= 4.5\]
It is positive! So now we know that above the critical point (1/3) the function is increasing. You can verify by checking the attached graph of the original function.

Answer 7

Thank you!