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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. >.

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4  
Find the domain of your function first... –  David Mitra Jun 4 '12 at 22:41
2  
dang it! Ln's can't be negative! Thanks lol –  ninja08 Jun 4 '12 at 22:42
    
Don't worry about $x \le -6$, $\log(x+6)$ is not defined. –  André Nicolas Jun 4 '12 at 22:42
    
@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. –  Hans Lundmark Jun 5 '12 at 6:44

3 Answers 3

The domain of $f(x)=\ln (x+6)-2$ is $x>-6$

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If this is already mentioned in the comments, why make this an answer? –  Joe Jun 4 '12 at 22:47
4  
Because the comments are not meant for answering the question, they should be used for clarifying it. –  Ben Millwood Jun 4 '12 at 23:22

The function does not exist if $x\le-6$ because $\operatorname{ln} (x)$ must take a positive value of $x$.

So using this condition,

$f'(x)=1/(x+6)$

$x \gt -6$ because the function does not exist and thus is not differentiable on $x\le-6$.

Therefore, $x+6\gt0$, so $1/(x+6) > 0$.

The derivative is strictly positive; thus, the function is always increasing.

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The function $\ln(x+6)$ is the inverse function of the strictly increasing function $g(x)=e^x-6$.

(We have $y=\ln(x+6)$ if and only if $e^y=x+6$ if and only if $x=e^y-6$.)

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