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- anonymous

f(x)=1/(4+9x^2) - sketch the region bounded by f, the x-axis, and the line x=1 and x=2

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- anonymous

f(x)=1/(4+9x^2) - sketch the region bounded by f, the x-axis, and the line x=1 and x=2

- schrodinger

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- julie

hmmm, well, I think basically this is a second derivative function. Paul has some pretty good notes here on how to use the second derivative to glean more about the shape of a function
http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII.aspx

- julie

Right now I'm watching this video:
http://www.brightstorm.com/math/calculus/applications-of-the-derivative/curve-sketching-with-derivatives

- julie

seems to explain it well also. but yeah you need information from both the first and second derivatives in order to correctly sketch the region

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- julie

been years since I've done so :p

- julie

$$f(x)=1/(4+9x ^{2})$$

- julie

^^ function using the equation editor thingy

- anonymous

Thanks Julie I think that helps me some!

- julie

hmmm, honestly going to have to look up the chain rule again lol

- julie

to even get the first derivative

- julie

hmmm. does anyone remember the chain rule? (looking)

- shadowfiend

So we have:
$$\frac{1}{4 + 9x^2}$$
Which makes:
$$(4 + 9x^2)^{-1}$$
We can apply the chain rule to this by first setting:
$$ u = (4 + 9x^2) $$
And:
$$\frac{d}{du} u^{-1} = -u^{-2}$$

- shadowfiend

But the chain rule says that taking the derivative of u when u consists of a function of x means we have to multiply by the derivative of the inner function:
$$ \frac{d}{dx} \left(4 + 9x^2\right) = \frac{d}{dx}4 + \frac{d}{dx}9x^2 $$
We can apply basic power rule and such and get:
$$ \frac{d}{dx} \left(4 + 9x^2\right) = 0 + 18x = 18x $$
And our end result is:
$$ -\left(4 + 9x^2\right)^-2 \cdot 18x = -\frac{18x}{\left(4 + 9x^2\right)^2} $$

- shadowfiend

That's the first derivative, you can apply the same reasoning to get the second derivative, and then use julie's advice to find the shape of the function.

- shadowfiend

(Feel free to correct me anyone -- I'm doing this in my head and haven't gotten a chance to double-check myself.)

- anonymous

One way of remembering the Chain Rule in this type of case is to remember that it will be the derivative of the outside function with the inside function left alone times the derivative of the inside function.

- anonymous

So, in this case once you write the fucntion as,
$$ (4+9x^{2})^{-1}$$
the "outside function" is the outermost function, i.e. the exponent of "-1" while the "inside function" is then $$4+9x^{2}$$, i.e. the stuff on the inside.

- anonymous

We know that the derivative of say $u^{-1}$ is $-u^{-2}$ so differentiating the outside, while leaving the inside alone would give,
$$-\left( 4+9x^{2}\right)^{-2}$$
Then multiply all this by the derivative of the inside, $18x$.
Or, putting it all together you get what shadofiend got.

- julie

wow thanks MD! very helpful. chain rule is a bit... confusing to decipher without explanation

- anonymous

Yes, it's can be difficult when trying to learn it (or scape the rust off skills long unused :) ). If you recall the outside/inside function way of using the chain rule it can help with a lot of the "simpler" problems.
Although I suspect "simpler" is in the eye of the beholder... :)

- anonymous

okay my text is working again

- shadowfiend

@MD Thanks -- by the way, for inline LaTeX, I'm afraid you have to use \ ( and \ ) (without the spaces) so that we don't start messing up when people start using $ for money :) So \(u^{-1}\) is \(u^{-2}\) should work.

- anonymous

Ah, I was wondering about that. It's been ages since I've seriously used LaTeX and I was thinking I'd misremembered how to do it (got a lot of rust there :) ). I'd forgotten about the other way of doing inline LaTeX. Thanks for reminding me!
On a side note a preview button might be good when typing into the box directly, i.e without using the Equation editor....

- shadowfiend

Yep. Ideally we'd like the equation editor itself to allow inline equations.

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