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In my question I am told to determine if the concavity is upwards or down words using this function

$$\frac{x^2+3}{x^2-4}$$

My problem is when i get to the second derivative I end up with $$-14x*2*2x-\frac{(x^2-4)*14}{(x^2-4)^3}$$

using derivative calculators it keeps showing $-56x^2$ at the top but with what I got the top does not turn into $56x^2$. I feel like I am combining the values the wrong way but I have tried to combine them in different ways and still do not get $56x^2$

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  • $\begingroup$ Note that $14x\cdot2\cdot2x=56x^2$. $\endgroup$ – Henning Makholm Jun 20 '15 at 17:27
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Note that $f(x) = \dfrac{x^2+3}{x^2-4} = 1 + \dfrac{7}{x^2-4}$. That makes it a lot easier to do the derivatives:

$$f'(x) = -7(x^2-4)^{-2}\cdot2x = -14x(x^2-4)^{-2}$$ $$\begin{split}f''(x) &= -14(x^2-4)^{-2} - 14x\cdot(-2)(x^2-4)^{-3}\cdot(2x) \\ &=-14(x^2-4)^{-2} + 56x^2(x^2-4)^{-3} \\ &=(42x^2+56)(x^2-4)^{-3}\end{split}$$

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  • $\begingroup$ how are you making it to where the 1 is on the outside and the 7 at the top $\endgroup$ – user249526 Jun 20 '15 at 21:23
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It is much easier to split this into partial fractions first, if you are going to take multiple derivatives.

$$f(x)=\frac {x^2+3}{x^2-4}=1-\frac 7{(x+2)(x-2)}=1-\frac 7{4(x-2)}+\frac 7{4(x+2)}$$

The first derivative is then $$f'(x)=\frac 74\left(\frac 1{(x-2)^2}-\frac 1{(x+2)^2}\right)$$ and $$f''(x)=-\frac 72\left(\frac 1{(x-2)^3}-\frac 1{(x+2)^3}\right)$$

And the sign of the second derivative clearly depends on comparing the two fractions. They can also be combined if this is found useful.

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