f(x)=((6x-3)/(x+6)) how do you find the intervals of decreasing and increasing...I found that there were no critical points yet I was also correct that the function increased (-I,-6) and (-6,I). I was wrong that there were no decreasing intervals though.

If f�Œ(x) > 0 at each point in an interval I, then the function is said to be increasing on I.

f�Œ(x) < 0 at each point in an interval I, then the function is said to be decreasing on I.

In your case :

f�Œ(x) = 39 /( x + 6 ) ^ 2

39 positive

( x + 6 ) ^ 2 always positive except
when x = - 6

In point x = - 6 function has vertical asymptote

So function :

f( x ) = ( 6 x - 3 ) / ( x + 6)

always increasing

P.S.

If you don't know how to find first derivation

Go on:

wolframalpha dot com

When page be open in rectangle type:

derivative (6x-3)/(x+6)

and click option =

When you see result click option:

Show steps

If you want to see graph of your function in google type:

function graphs online

When you see list of results click on:

rechneronline.de/function-graphs

When page be open in blue rectacangle type:

(6x-3)/(x+6)

Set :

Range x-axis from -100 to 100

Range y-axis from -100 to 100

Then click option :

Draw

You will see graph of your function.

To find the intervals of increasing and decreasing for a given function, you need to analyze the behavior of the first derivative of the function. In this case, we need to find the derivative of f(x):

f(x) = (6x - 3)/(x + 6)

To find the derivative, we can use the quotient rule:

f'(x) = [(6)(x + 6) - (6x - 3)(1)] / (x + 6)^2

Now let's simplify the above expression:

f'(x) = (6x + 36 - 6x + 3) / (x + 6)^2
= 39 / (x + 6)^2

Since the derivative is a constant (39) divided by a squared quantity (x + 6)^2, the sign of the derivative will not change. Therefore, the function f(x) is either always increasing or always decreasing.

To determine which is the case, we can evaluate the derivative at an arbitrary point, such as x = 0:

f'(0) = 39 / (0 + 6)^2
= 39 / 36
= 1.083...

Since f'(0) is positive, it indicates that f(x) is increasing on the entire domain. This means there are no intervals of decreasing for the given function.

Therefore, it seems there might be some misunderstanding or mistake in your analysis. If you provide more information or steps you followed, we can identify the error and correct it.