Point of Inflection (second derivative)- easy

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June 6th 2010, 04:20 PM
dkssudgktpdy

Point of Inflection (second derivative)- easy

Can anyone solve this one?:confused:
I got stuck..

f(x) = 4/ (x^2 + 8)

I need f''(x) second derivative for this
June 6th 2010, 04:21 PM
mr fantastic

Quote:

Originally Posted by dkssudgktpdy

Can anyone solve this one?:confused:
I got stuck..

f(x) = 4/ (x^2 + 8)

I need f''(x) second derivative for this

Can you get the first derivative? Please show all your work.
June 6th 2010, 04:26 PM
dkssudgktpdy

Quote:

Originally Posted by mr fantastic

Can you get the first derivative? Please show all your work.

yup.
f(x) = 4 (x^2 +8)^-1

= -4 (x^2 +8)^-2 (2x)

= -8x (x^2 +8)^-2
f''(x)= 16x (x^2 +8)^-3 (2x)
= 32x (x^2 +8)^-3
0 = 32x/ (x^2 +8)^3
x=0

and.. 0 is not the right answer. Where did I make a mistake?
June 6th 2010, 04:39 PM
TheEmptySet

Quote:

Originally Posted by dkssudgktpdy

yup.
f(x) = 4 (x^2 +8)^-1

= -4 (x^2 +8)^-2 (2x)

= -8x (x^2 +8)^-2
f''(x)= 16x (x^2 +8)^-3 (2x)
= 32x (x^2 +8)^-3
0 = 32x/ (x^2 +8)^3
x=0

and.. 0 is not the right answer. Where did I make a mistake?

You need to use the product rule or quotient.

$\frac{d^2y}{dx^2}=\frac{(x^2+8)(-8)-(-8x)(2(x^2+8)(2x))}{(x^2+8)^4}$
June 6th 2010, 05:14 PM
dkssudgktpdy

Quote:

Originally Posted by TheEmptySet

You need to use the product rule or quotient.

$\frac{d^2y}{dx^2}=\frac{(x^2+8)(-8)-(-8x)(2(x^2+8)(2x))}{(x^2+8)^4}$

I got it.
Thank you!

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