Question

MEDALS!!
Need help determining where these functions increase/decrease as well as where the horizontal and oblique asymptotes are located. Thanks!!

Answer 1

holy crap those look hard

Answer 2

To find where f(x) is increasing, find the first derivative, f'(x) and solve the inequality f'(x) > 0.

Answer 3

for 2x+5/8x+9 i believe the critical point is a -9/8. would the function be increasing at (-9/8,inf) and decreasing at (-inf,-9/8)? I graphed this as well and the concavity seems to align with everything i just mentioned..idk..

https://www.desmos.com/calculator/7d29vxyqxu

Answer 4

post one problem at a time. You'll get a higher chance of getting help

Answer 5

sorry, im rather new :/

Answer 6

f'(x) = -22 / (8x+9)^2
Yes, f'(x) (as well as f(x)) is undefined at x = -9/8.
Critical point is where f'(x) is 0 or undefined and therefore, x = -9/8 is a critical point.
In f'(x) we notice, the bottom is always >=0 and the top is always negative.
Therefore, f'(x) is always negative which means f(x) is always decreasing (except at x = -9/8 where f(x) is undefined and therefore excluded from the domain).

So the function is decreasing in the interval: (-infinity, -9/8) U (-9/8, infinity)

Answer 7

And you will find your horizontal asymptote by seeing what this function will equal with \(\lim_{x \rightarrow \infty } \)

Answer 8

f(x) = (2x+5) / (8x + 9) = (2 + 5/x) / (8 + 9/x)
Take the limit of f(x) as x->infinity. 5/x and 9/x will ->0 and you are left with y = 2/8 = 1/4. So horizontal asymptote is y = 1/4.

Answer 9

Ok so i finished 2x+5/8x+9 . yeah i that's the horizontal asymptote that i got. my last question pertains to x^3/((x^2)-4), how to i write out the increasing and decreasing functions for seemingly 3 curves? https://www.desmos.com/calculator

Answer 10

https://www.desmos.com/calculator/fcv0uqnp3q

Answer 11

I know you're supposed to use brackets for 0 when its increasing (-2,0] but decreasing at [0,-2) however, how do i incorporate that with the other curves on either side?

Answer 12

The function is undefined at x = -2 and x = +2 and should be excluded from the domain.
The next step is to find all the critical points:
Find the derivative of f(x), equate it to 0 and solve for x.
Then we can work on find all the intervals where the function is increasing / decreasing.

Answer 13

ok i got 0, +/- 2sqrt(3)

Answer 14

Yes, good. Combining the critical points and the points where the function is not defined, we have the following intervals to consider:
-infinity -2sqrt(3) -2 0 +2 +2sqrt(3) +infinity

We can clearly see from the graph or by picking a suitable x value in each interval and finding whether f'(x) is positive or negative at those points that:
f(x) is increasing in: (-infinity, -2sqrt(3)) U (2sqrt(3), infinity)

Answer 15

f(x) is decreasing in the intervals: (-2sqrt(3), -2), (-2, +2), (+2, +2sqrt(3))
You can either use the comma to separate the intervals or use the union symbol to combine the intervals depending on what was taught to you.

Answer 16

Thanks so much for your help! I had to add a bracket to every 2sqrt(3) but everything else looks great. My final question would be how do I find the oblique asymptote to x^3/((x^2)-4)? I tried long division and ended up getting x+1, its not correct :/

Answer 17

Yeah, some textbooks include the maxima/minima points in the increasing/decreasing intervals and use '[ or ]' other text books exclude them and use '( )'. I prefer the latter but it looks like your course uses the former. Those maxima, minima points are also referred to as the "stationary points" -- stationary because the function is neither increasing nor decreasing at those points and therefore should be excluded from the increasing / decreasing intervals denoted by '( or )'.

Answer 18

To find the oblique/slant asymptote you do the long division and express the function as a quotient and remainder. What do you get after the division?

Answer 19

(x+1)+(4x+4/((x^2)-4))

Answer 20

If you combine the above terms under a common denominator you get x^3 + x^2 in the numerator.

Answer 21

lol forgot that last bit. again..thanks!

Answer 22

x^3 / (x^2 - 4) = { x(x^2 - 4) + 4x } / (x^2 - 4)
= x(x^2 - 4) / (x^2 - 4) + 4x / (x^2 - 4)
f(x) = x + 4x / (x^2 - 4)
As x ->infinity, 4x / (x^2 - 4) -> 0
Therefore, y = x is the oblique asymptote.

In the desmos graphing website, try y = x in the line below the previous function and see if it is indeed a slant asymptote to the function.

Answer 23

yeah the slope is 1

Answer 24

and both functions only approach the slant

Answer 25

Yes, slope is 1 and also should pass through the origin and hence y = x is the slant asymptote.
Your answer which was y = x + 1 also has a slope of 1 but it has a y-intercept of 1 and if you plot it you will notice it intersects the curve and so it is not a slant asymptote.

Answer 26

i see

Answer 27

In your answer to 5, put the lower interval first and then the higher interval.

Answer 28

5) (-2, 0) U (2, inf)