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I am trying to graph $$f(x) = \frac{x^2-4}{x^2+4}$$

It seems pretty simple to me but I can't finish it correctly.

I know that there is a horizontal asymptote at $1$ beacuse the degrees on the variable at the same and $\frac{x}{x}$ is 1.

I know that it is a negative function until zero, and then positive because the only critical number of the derivative $$\frac{16x}{(x^2+4)^2}$$ is going to be zero since the denominator can't be zero and the only root of the top is $0$.

This then tells me that there is no local max but only a minimum which is at $0$ which gives me $-1$.

Trying to find concavity $$\frac{16(x^2+4)^2 - 16x (4x(x^2+4))}{(x^2+4)^4}= \frac {-64x^5 + 16x^4 + 256x^3 + 128x^2 + 256}{(x^2+4)^4}$$ Which I have no idea how to really work with I can't see to get anything workable out of that.

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    Have you considered the zeros of $f(x)$?2012-04-06
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    Or that $f(x)=f(-x)$?2012-04-06
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    I know that the y interecpt is -1 and that the x is 2 and maybe -2 but I don't quite understand why.2012-04-06
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    Set $x=2$ or $x=-2$ and see what happens. There are some things you say you know which are not true. Rather than starting with the properties (e.g. zeros) of multiple derivatives, it is worth thinking of the zeros and symmetries of the function itself. When would the value of the function be greater than 1 or less than -1?2012-04-06
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    I am not sure but I know that asymptotes can sometimes be ignored so I try not to consider them too much, I don't quite have all the graph stuff memorized since I haven't used it in over a year.2012-04-06
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    I think that you will be required to do a little work even to graph a function like this one. Noticing what should be obvious does help. It should not be a question of what is in your memory, but rather one of the tools you have to approach the problem. Zeros, asymptotes, turning points, maximum and minimum values - all are very basic. Looking for the easiest ones first comes with practice.2012-04-06
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    I don't really care about learning math anymore, I just need a GPA to do something with my life so learning is no longer important. I have a test on monday and I have to get answers correct to get maximum points on the test, to do this I just have to list the increase and decrease, zeros, min/max and concavity.2012-04-06
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    Your second derivative has an incorrect sign (you should have $x^2+4$ in the second summand, not $x^2-4$); factor out $x^2+4$ in the numerator and cancel it with one of the factors in the denominator before simplifyin. Then simplify some more.2012-04-06
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    The GPA without the knowledge it supposedly represents is worthless.2012-04-06
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    @ArturoMagidin +1. You do indeed have a sign error and can simplify it a bit.2012-04-06
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    That was an error in typing it up, my work on paper does not have that error. I disagree about the GPA, the only thing colleges actually care about is a GPA, if the GPA is not good enough you do not matter as a person to the college.2012-04-06
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    Shouldn't *you* care though about the knowledge?2012-04-06
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    *Colleges* care about GPA; but unless you plan to be a career student, the GPA will not matter later. A good GPA without the knowledge it supposedly represent may fool someone into hiring you, but won't keep them from firing you if you don't exhibit the knowledge that is needed.2012-04-06
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    Not anymore, I did that before and I ended up getting bad grades because I spent too much time trying to just learn how things work and why and not spending enough time memorizing what will be on the next test.2012-04-06
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    And so we're back to passive-aggressive Jordan. Do as you wish.2012-04-06
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    I don't see how this is being passive-aggressive. I just found out that I am wasting my time in school basically because my grades will never be good enough to ever transfer. So yes, GPA matters a lot. It is basically all that matters.2012-04-06
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    It is not "negative until zero". It's negative at $0$ and changes signs at $2$ and $-2$. It's an _even_ function, so it can't change signs at $0$.2012-04-06
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    I don't understand, why can't an even function change at zero? Doesn't x^2?2012-04-06
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    No, $x^2$ doesn't change signs at $0$: it's positive on both sides of $0$. It can't change signs at $0$ because it has to take the same values to the left of $0$ as it takes to the right of $0$. There is *symmetry* about the $x$-axis.2012-04-07

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Your second derivative looks wrong. You should get:

$$\frac{64-48x^{2}}{(x^2+4)^{2}}$$

If you would mind showing your work on taking the derivative, that would be helpful. Did you use quotient rule or rewrite and use product rule?

From that, you can test a few points to see where the graph is concave up or concave down.

enter image description here

The graph should look something like that (to give you an idea for test interval points if you aren't aware of what to note when doing the second derivative test). The picture was taken from Wolfram.

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    I am almost certain my derivative is correct since that is what wolfram gives me. Just to keep it realistic I had wolfram multiply out everything like I was entering in the quotient rule by hand. That way it wouldn't reduce it so much. It matched what I did on my own.2012-04-06
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    Are you sure you typed it in correctly? Check http://www.wolframalpha.com/input/?i=second+derivative+of+%28x%5E2-4%29%2F%28x%5E2%2B4%29.2012-04-06
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    The second derivative is $16\frac{4-3x^2}{(4+x^2)^3}$. Don't really care about Alpha's opinion on the matter.2012-04-06
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This particular function and its derivatives can benefit from using several perspectives: $$ f(x)=\frac{x^2-4}{x^2+4}=\frac{(x+2)(x-2)}{x^2+4}=1-8(x^2+4)^{-1} $$ $$ f\,'(x)=16x(x^2+4)^{-2} $$ $$ f\,''(x)=16(x^2+4)^{-3}(4-3x^2)=-16\frac{3x^2-4}{x^2+4}=-48\frac{\left(x+\frac2{\sqrt3}\right)\left(x-\frac2{\sqrt3}\right)}{x^2+4} $$ Since $x^2+4$ is always positive, the factored forms are workable -- they give you all the critical points and allow you to draw number lines below. $$ \matrix{ x & -\infty & -2 & -\tfrac2{\sqrt3} & 0 & \tfrac2{\sqrt3} & 2 & \infty \\ f &+&0&-&-&-&0&+\\ f' &-&-&-&0&+&+&+\\ f'' &-&-&0&+&0&-&-\\ } $$ The columns for $\pm\infty$ are meant to represent the signs for $x$ sufficiently large, i.e. past the last finite critical point of $f$ or one of its derivatives; for example, the $-48$ above tells us $f\,''$ should be negative for $x$ larger than all of its roots.

From this number line, you can infer the graphical behavior and make a rough sketch.

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Correct second derivative: $$\begin{align*} f(x) &= \frac{x^2-4}{x^2+4}\\ f'(x) &= \frac{(x^2+4)(x^2-4)' - (x^2-4)(x^2+4)'}{(x^2+4)^2}\\ &= \frac{(x^2+4)(2x) - (x^2-4)(2x)}{(x^2+4)^2}\\ &= \frac{2x(x^2+4-x^2+4)}{(x^2+4)^2}\\ &= \frac{16x}{(x^2+4)^2}.\\ f''(x) &= \frac{(x^2+4)^2(16x)' - 16x\Bigl( (x^2+4)^2\Bigr)'}{\Bigl((x^2+4)^2\Bigr)^2}\\ &= \frac{16(x^2+4)^2 - 16x(2(x^2+4)(x^2+4)')}{(x^2+4)^4}\\ &= \frac{16(x^2+4)^2 - 16x(2)(x^2+4)(2x)}{(x^2+4)^4}\\ &= \frac{16(x^2+4)\Bigl( x^2+4 - 4x^2\Bigr)}{(x^2+4)^4}\\ &= \frac{16(4-3x^2)}{(x^2+4)^3}. \end{align*}$$ The denominator is never zero; the numerator is zero exactly when $4-3x^2=0$. This is a parabola that opens down, so it will be positive between the roots and negative before and after both roots.