Graph Solutions for y=1/(3+x^2)

[f22archrer]

Homework Statement

y = 1/ (3+x^2))

Homework Equations

y int; y =1/3 (((0,1/3)
HA:
lim x- (1/x^2) / (3/x^2+ x^2+x^2))
y =0
VA: none
HA


x_ infine x__ + infinite { next two lines are under it, for x infinite and x + infinite)

TV -1000 1000
y= 0 0

above above

f '(x) =[ (3+x^2)(0) - 1(2x) ] /[ ((3+x^2)^2 ]
= 2x / (3+x^2)^2

fx = 0 at x = 0

what else am i missing?? to draw a complete graph : teacher said a lot was missing and try again ,but i don't see anything missing!

The Attempt at a Solution

 

[symbolipoint]

y = 1/ (3+x^2))
$y=\frac{1}{3+x^2}$

Before even using derivatives, see that because the denominator is always positive, y is always positive, above the x axis. x can be zero, rendering the largest possible value for y. As x becomes either more negative without bound or more positive without bound, the rational expression moves closer to zero but will never reach zero.

You should find a maximum at x=0 and asymptote y=0.
Derivatives may give you a bit other information, like inflection points if you find second derivative... you should study how. The first derivative should be equal to zero at either a maximum or a minimum; and in the example given, you should expect to find a maximum (at x=0).

 

[SammyS]

Homework Statement

y = 1/ (3+x^2))

It helps us to help you, if you give a complete statement of the problem.

Let's see ...

You've given us a title, "curve sketching" , an equation giving y as a function of x, then below (at the very end), you mention that the work you have shown here "has a lot missing" according to your teacher..

I guess you are supposed to use calculus and some algebra skills to determine all you can regarding the given function so that you can adequately sketch a graph of the function.

Homework Equations

y int; y =1/3 (((0,1/3)
HA:
lim x- (1/x^2) / (3/x^2+ x^2/x^2))
y =0
VA: none

HA

x_ infinite x__ + infinite { next two lines are under it, for x infinite and x + infinite)

TV -1000 1000
y= 0 0

above above

It took me quite a while to figure out what you were trying to do in the above snippet.

It looks like you're trying to find a bit more than simply the HA for x → ± ∞ , by using some fairly large test points.

f '(x) =[ (3+x^2)(0) - 1(2x) ] /[ ((3+x^2)^2 ]
= 2x / (3+x^2)^2

fx = 0 at x = 0

what else am i missing?? to draw a complete graph : teacher said a lot was missing and try again ,but i don't see anything missing!

The Attempt at a Solution

You have a sign error in your first derivative.

$\displaystyle f'(x)=\frac{-2x}{(3+x^2)^2}$​

Where is f '(x) > 0 ? Where is f '(x) < 0 ?

What does it mean that f '(0) = 0 ?

What is f ''(x) ? Where is it +, 0, - ?

Are f(x) f '(x), f ''(x) even or odd respectively?

 

[symbolipoint]

REALLY BIG HINT: The numerator of the simplified form of the second derivative is
-2(3+x²)+8x²

Where is f''(x)=0?

Another hint, just as important: Where is f'(x)=0 and what does this mean?

 

[f22archrer]

i guess i have done a lot wrong from the start!

can you tell me exactly the steps i need to start from the beginning using calculus.

all i know is to find y intercept, horizontal asymptote, V.A , using the test values and finding how the graph will look...

what else! please help

 

[SammyS]

i guess i have done a lot wrong from the start!

can you tell me exactly the steps i need to start from the beginning using calculus.

all i know is to find y intercept, horizontal asymptote, V.A , using the test values and finding how the graph will look...

what else! please help

It's not that you did all that much wrong. You had a sign error in the first derivative.

Mainly, as your teacher said, you could have gone much further.

You found the y-intercept, but should have mentioned that there's no x-intercept.

Did you notice that your function is even?

You found the Horizontal Asymptote and also noted that there's no Vertical Asymptote. At this point you might observe that the domain of the function is (-∞, +∞).

Determining that the function approaches the HA from above was fine. It just took me some time to figure out what you were doing how you were doing it.

The first derivative is $\displaystyle \ y'=-\frac{2x}{(3+x^2)^2}\ .$ You can use that to find the min/max for the function, and also to find where the function is increasing/decreasing as y' is positive/negative.

Use the second derivative to find any inflection points and to determine the concavity of the function.

Somewhere along the line, you could also determine the range of the function.

 

[f22archrer]

thanks , that helped me get back on track.