# Math Help - Derivative question

1. ## Derivative question

$f(x)=\frac{2x^2+4x+2}{x^2+1}$

$\lim_{x\ \rightarrow\ \infty}f(x)=$

(I am not familiar with this notation at all... )

$f'(x)=$

$f'(x)=0$ when $x=A,\:B$

$A=\:\:,\:B=$

$\int_A^Bf(x)dx=$

I'm not even sure where to start. I mean, I can factorise the equation, but would that help? I know that the derivative of the equation is the slope at any given value of $x$, and the slope is 0 for each minima/maxima. But I don't know how to answer these questions...

2. Originally Posted by davidman
$f(x)=\frac{2x^2+4x+2}{x^2+1}$

$\lim_{x\ \rightarrow\ \infty}f(x)=$

The limit as x goes to infinity of the function said above.

How to do this problem? Divide every term by $x^{2}$

$\lim_{x\ \rightarrow\ \infty} f(x)=\frac{\frac{2x^2}{x^2}+\frac{4x}{x^2}+\frac{2 }{x^2}}{\frac{x^2}{x^2}+\frac{1}{x^2}}$

Do you see how to solve the problem? Remember that as x goes to infinity, denominators become INFINITELY large ... so what does that do to numbers/fractions? It makes them zero.

Buuuuuuuuut ... don't forget to do any necessary "cancellations."

3. Originally Posted by davidman
$f(x)=\frac{2x^2+4x+2}{x^2+1}$

$\lim_{x\ \rightarrow\ \infty}f(x)=$

(I am not familiar with this notation at all... )

$f'(x)=$

$f'(x)=0$ when $x=A,\:B$

$A=\:\:,\:B=$

$\int_A^Bf(x)dx=$

I'm not even sure where to start. I mean, I can factorise the equation, but would that help? I know that the derivative of the equation is the slope at any given value of $x$, and the slope is 0 for each minima/maxima. But I don't know how to answer these questions...
If the only thing you know about the derivative is that it is the slope (strictly speaking it is the slope of the tangent line to the graph- only lines have "slope".) then these questions are way over your head. Go to your teacher and make sure he/she knows your difficulty.

4. Originally Posted by Lord Darkin
Divide every term by $x^{2}$
Thanks... so

$\lim_{x \rightarrow\ \infty}f(x)=\lim_{x \rightarrow\ \infty}\frac{2+\frac{4}{x}+\frac{2}{x^2}}{1+\frac{ 1}{x^2}}=2$

Although, I'm not entirely sure why to divide every term by $x^2$ specifically... I mean, I can see that it's the largest exponent of $x$. Is that how you decide what to divide by?

Originally Posted by HallsofIvy
This is true, but I actually understated my ability/experience with derivatives a bit.

I don't remember much at all of what I studied (several years ago), but I think for instance if $F(x)=2x^2+4x+2\:\:,\:\:F'(x)=4x+4$

How to get the derivative of $f(x)=\frac{2x^2+4x+2}{x^2+1}$ I forgot.

And lastly, unfortunately, I have no teacher, just books and the internet. Really grateful for all the help though.

5. Yes, I divide by the largest exponent of x. That's how I decide.

6. Then you had better keep reading those books! Hopefully, just after telling you that the derivative is the slope of the tangent line, they will have a bunch of "rules" for differentiating.

For all of these problems, it helps a lot to recognise that
$\frac{2x^2+ 4x+2}{x^2+ 1}= \frac{2x+2}{x^2+ 1}+ \frac{4x}{x^2+ 1}= 2+ \frac{4x}{x^2+1}$. Since the denominator in that fraction has a higher power than the numerator, it goes to 0 as x goes to infinity.

To differentiate that, use the quotient rule. Do you know what that is?

For the second question, set the derivative equal to 0 and solve for x. Since the derivative is also a fraction, the first thing you can do is multiply both sides of the equation by the denominator.

Integrating, use the substitution $u= x^2+ 1$ for the fraction. I would hope you would not have any problem integrating "2" but integration, in most books, comes well after the definition of the derivative!

7. Thank you!

$f(x)=\frac{2x^2+4x+2}{x^2+1}$

$f'(x)=\frac{(2x^2+4x+2)'(x^2+1)-(2x^2+4x+2)(x^2+1)'}{(x^2+1)^2}$

$=\frac{(4x+4)(x^2+1)-2x(2x^2+4x+2)}{(x^2+1)^2}$

$=\frac{4x^3+4x+4x^2+4-4x^3-8x^2-4x}{(x^2+1)^2}$

$=\frac{-4x^2+4}{(x^2+1)^2}$

$=\frac{-4(x^2-1^2)}{(x^2+1)^2}$

$=\frac{-4(x+1)(x-1)}{(x^2+1)^2}$

$f'(x)=0\:\:,\:\:x=-1\:\:,\:\:1$

I notice that the denominator isn't going to become equal to zero if the $x$ there is squared.

Anyways, this gives me the limits of the integration to be done. Or whatever it's called... xD Next step,

$\int_{-1}^1f(x)dx=$

As pointed out; $f(x)=2+\frac{4x}{x^2+1}=2+\frac{4x}{u}$

$\int_{-1}^1f(x)dx=\int_{-1}^1\left(2+\frac{4x}{u}\right)dx$

$\left[2x\right]_{-1}^1+\int_{-1}^1\left(\frac{4x}{u}\right)dx=$

and that is where I get stuck

thanks for all the help. Not quite sure how to do the following step(s), but I'll try and look it up.

Edit: found a formula that looked relevant $\int{af(x)}dx=a\int{f(x)}dx$

$=\left[2x\right]_{-1}^1+4\int_{-1}^1\left(\frac{x}{u}\right)dx$

Edit 2: ok, I think I got to the answer in my book somehow

another formula $\int_a^bf(x)dx=F(b)-F(a)$ inspired me to try and plug in the two limits and simplify, but I'm not entirely sure why it works.

$\left[2x\right]_{-1}^1+4\int_{-1}^1\left(\frac{x}{x^2+1}\right)dx=\:$ $\color{red}2(1)-2(-1)+4[\frac{1}{1^2+1}-\frac{-1}{-1^2+1}]$

$=4(\frac{1}{2}+\frac{1}{2})=4$

$\int_{-1}^1f(x)dx=4$

What I'm really wondering now is why I can stop simplifying when I have $\frac{x}{x^2+1}$

And is it really necessary to move the $4$ outside the $\int$?

Edit 3: Noticed my calculation was off... $2(1)-2(-1)=4\neq0$

Ok, seems I'm only going to bury myself deeper if I keep editing here. Anyone care to help me out?

Edit 4: Ok, last edit

didn't realise I was posting this in the wrong subforum. Sorry about that. Anyways, I tried graphing the bit I couldn't find the integral for. It's pretty obvious why the integral of that on a symmetrical space across the origin would be equal to zero... so I guess I'll keep that in mind when I can't figure out the integral.

So thanks again everyone.