Question

Find where the Function is increasing and decreasing:

Answer 1

\[f(x)= \frac{ x^2 }{ x^2-16 }\]

Answer 2

By graphing or using calculus?

Answer 3

I got it's Increasing from (-infinity,-4) and decreasing from (4, infinity).

Answer 4

Calculus.

Answer 5

I also put It's increasing from (-infinity,0) and decreasing from (0,infinity) which was incorrect.

Answer 6

For the derrivative use the quotient rule:
f'(x) = [ [2x](x^2 - 16) - (x^2)[2x] ] / [x^2 - 16]^2

It's increasing from -infinity to -4(where it has a vertical asymptote) and then from -4 to 0

Answer 7

That's what I put but it was incorrect.

Answer 8

hmm

Answer 9

Hmm let me try something...

Answer 10

I should add:

It's increasing from -infinity to -4(where it has a vertical asymptote) and increasing from -4 to 0

AND

It's decreasing from 0 to 4(where it has another vertical asymptote) and then decreasing again from 4 to infinity

Answer 11

So like Increasing when: (-infinity,-4)U(-4,0) ?

Answer 12

That's probably my mistake. I included the -4 in my answer.

Answer 13

you should probably separate those with a comma and not use union

Answer 14

I have to use interval notation.

Answer 15

the function is not increasing on (-infinity,-4)U(-4,0)


it is increasing on (-infinity,-4) and it is increasing on (-4,0)

Answer 16

Why do I not use Union?

Answer 17

by definition...

a function \(f\) is increasing on a set \(A\) if for any \(x,y\in A\) with \(x<y\) then \(f(x)<f(y)\)

Answer 18

I see.

Answer 19

if you use (-infinity,-4)U(-4,0) then we could pick x=-5 and y=-3

is f(-5)<f(-3)?

Answer 20

Nope.

Answer 21

usually these problems are stated...find the larges open intervals on which the function is increasing / decreasing

Answer 22

*largest

Answer 23

Here I should specify the question. One moment.

Answer 24

they probably want union then...though that is incorrect

Answer 25

I know. Stupid web-assign :P .

Answer 26

(-4,4)-----decreasing
R-(-4,4)-----increasing

Answer 27

yes I am aware of the discontinuity...that is what is causing the issue.


f(-5) is bigger that f(-3) so it is not increasing on (-infinity,-4)U(-,4,0)

though it is increasing on the two individual sets

Answer 28

Yes! Union gives the correct answer :) . Thanks everyone :) .

Answer 29

I agree there should be a comma though...

Answer 30

if they asked where the derivative was positive then the answer is (-infinity,-4)U(-,4,0)

Answer 31

I forgot there was an asymptote at -4 and 4.

Answer 32

@Zarkon but what about a function like this:

the function is greater on the right side than the left and is increasing but still has a discontinuity