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StuartYates
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#1
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Hi everyone!

Am I correct in saying that differentiating f(x)=x/x^2+2 gives

2-x^2/(x^2+2)^2

??
and how would I then use this to find the set of values of x for which f'(x)<0

Thank you!
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RichE
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(Original post by StuartYates)
Hi everyone!

Am I correct in saying that differentiating f(x)=x/x^2+2 gives

2-x^2/(x^2+2)^2

??
and how would I then use this to find the set of values of x for which f'(x)<0

Thank you!
Yes, and for the second part notice that the denominator is positive. So f'(x)<0 when 2-x^2 <0.
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Aitch
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(Original post by StuartYates)
Hi everyone!

Am I correct in saying that differentiating f(x)=x/x^2+2 gives

2-x^2/(x^2+2)^2

??
and how would I then use this to find the set of values of x for which f'(x)<0

Thank you!
If you need to double-check your differentiation, try this:

http://www.calc101.com/webMathematic...es.jsp#topdoit

Free for derivatives. If you want to do integrals,

http://www.calc101.com/webMathematica/integrals.jsp

you get the integration result free, but need to subscribe if you want all the steps.

Aitch
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StuartYates
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Report Thread starter 14 years ago
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That's fantastic thank to Aitch.

Very useful to check my answers!!
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Nima
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(Original post by StuartYates)
f(x) = x/(x^2 + 2).

1.) Find f ' (x)
2.) Find the set of values of x for which f ' (x) < 0
1.) Let y = f(x) --> y = x/(x^2 + 2)
Let u = x --> du/dx = 1
Let v = (x^2 + 2) --> dv/dx = 2x
Hence: f ' (x) = dy/dx = [v(du/dx) - u(dv/dx)]/(v^2) = [(x^2 + 2) - 2x^2]/(x^2 + 2)^2

--> f ' (x) = (2 - x^2)/[(x^2 + 2)^2]

2.) For f ' (x) < 0:
--> (2 - x^2)/[(x^2 + 2)^2] < 0
--> 2 - x^2 < 0
--> Critical Points: x = (+/-) Sqrt(2)
Let g(x) = 2 - x^2:
When x < -Sqrt(2): g(x) < 0 --> True
When -Sqrt(2) < x < Sqrt(2): g(x) > 0 --> False
When x > Sqrt(2): g(x) < 0 --> True

Hence for f ' (x) < 0:
x < -Sqrt(2), x > Sqrt(2)

Nima
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