6
$\begingroup$

What I'm doing is finding where this function is decreasing or increasing.

Here is the original function:

$f(x) = \ln(x+6)-2$

I take the prime when I believe is:

$f'(x)= \dfrac{1}{x+6}$

Then I made a sign chart.

I know right off the bat that there is nothing that make this function equal to zero because the numerator doesn't have an $x.$

The denominator can make the function undefined, and it's undefined at $-6.$ So thats the number I use on my sign chart.

I plugged the first value $-10$ into the prime function and it gives me a negative value:

$f'(-10)= \frac{1}{(-10+6)}$

$ = \frac{-1}{4}$

Then I plugged the $0$ in and I got

$f'(0)= 1/6$

It should look something like this:

   -          n | d         + 

-----(-10)------ ((-6)) -------(0)------

My homework is saying the function is never decreasing. >.

  • 4
    Find the domain of your function first...2012-06-04
  • 2
    dang it! Ln's can't be negative! Thanks lol2012-06-04
  • 0
    Don't worry about $x \le -6$, $\log(x+6)$ is not defined.2012-06-04
  • 0
    @ninja08: It's not "ln's" that can't be negative, it's the $x$'s that you plug into them that can't be negative.2012-06-05

3 Answers 3