Question

calculate the derivative of 4x/(x^2-36). And find the interval on which it is decreasing.
I know that the derivative is (4((x^2)-36)-8x^2)/((x^2)-36)^2
but I don't know how to find where it is decreasing since it doesn't have any critical numbers

Answer 1

I got the same thing for the derivative, one second on where it is decreasing

Answer 2

So there are critical numbers at -6 and 6 because they make the denominator 0 which means the fraction will either be 0 or nonexistent--either way they are critical numbers

Answer 3

You can also find this algebraically by doing the following:

0 = (4((x^2)-36)-8x^2)/((x^2)-36)^2

Multiply by denominator on both sides

0 = 4((x^2)-36)-8x^2

add 8x^2 both sides

8x^2=4((x^2)-36)

Divide by 4 both sides

2x^2=x^2 - 36

Subtract x^2 both sides

x^2 = 36

Square root both sides

x = 6 or x = -6

Answer 4

So then you can take one derivative from each of the following intervals to find where it's increasing:

x<-6

-6<x<6

6<x

Answer 5

So these are examples

f'(-7)=-340/169
f'(0)=-1/9
f'(7)=-340/169

So it is decreasing in all of those intervals because they are all negative derivatives. You can check your answer by looking at the graph of f(x) (attached).

Answer 6

In that graph you can see that the function is decreasing everywhere, so our answer makes sense!

Answer 7

thanks!

Answer 8

Hope I helped! Feel free to ask questions if you have any. Also, wolframalpha.com is a really great resource for performing calculations (not as great at explaining, though).