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Hi! I have a question concerning derivatives :)
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\[\frac{ d }{ dx }(4-x^{2}-\frac{ 16 }{ x^{2} })\]
Now, according to Wolfram|Alpha, the correct answer is\[x=\pm2\] I just don't know how to get there :P
Here's what I did:\[\frac{ df }{ dx }=-2x-16(\frac{ -2x }{ x^{4} })\]\[=-2x+\frac{ 32x }{ x^{4} }=-2x+\frac{ 32 }{ x^{3} }\]
btw, df/dx=0
Then I factor out the x, right?\[=x(-2+\frac{ 32 }{ x^{4} })=0\]
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I kinda know how to get to x=+/-2, I just don't know why I don't get x=0 as well :/
Look at the graph; there is a horizontal asymptote at x = 0.
It is not in the domain. I mean x =/= 0
Ooooh, of course! :P Thanks a lot! :P
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