Sketch Graph of f(x): Find Inflection Points

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SUMMARY

The discussion focuses on finding the inflection points of the function f(x) = (x^2)/((x-2)^2). The second derivative, f''(x), was derived as f''(x) = [ (x-2)^3 ] (-8) [ (3x^2) + x + 2 ] / [(x-2)^8]. A key challenge identified is the inability to factor the polynomial 3x^2 + x + 2, which is necessary for determining where f''(x) equals zero. A more efficient approach suggested involves simplifying the first derivative, f'(x), before taking the second derivative.

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Homework Statement



Sketch graph of f(x)= (x^2)/((x-2)^2). I have retrieved the first derivative, found the critical points, and also have the vertical asymptote. I seem to be having trouble trying to find the inflection points... I can't seem to find a nicely factored f''(x).

Homework Equations


The Attempt at a Solution



so far I have f''(x)= [ (x-2)^3 ] (-8) [ (3x^2) + x + 2 ] / [(x-2)^8]

I can't factor the 3x^2+x+2 to be able to find where f''(x)=0 and thus revealing the inflection points :(. Help? thankyou...
 
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sdoug041 said:

Homework Statement



Sketch graph of f(x)= (x^2)/((x-2)^2). I have retrieved the first derivative, found the critical points, and also have the vertical asymptote. I seem to be having trouble trying to find the inflection points... I can't seem to find a nicely factored f''(x).

Homework Equations





The Attempt at a Solution



so far I have f''(x)= [ (x-2)^3 ] (-8) [ (3x^2) + x + 2 ] / [(x-2)^8]

I can't factor the 3x^2+x+2 to be able to find where f''(x)=0 and thus revealing the inflection points :(. Help? thankyou...
For problems like these, it's more efficient to get the derivative in its simplest form before you take the derivative again.

For your function, I found this for f'(x):
[tex]f'(x)~=~\frac{2x(x - 2)^2 - 2x^2(x - 2)}{(x - 2)^4}[/tex]
By finding common factors in the numerator, I was able to simplify it in this way
[tex]f'(x)~=~\frac{2x(x - 2)(x - 2 - x)}{(x - 2)^4}~=~ \frac{-4x}{(x - 2)^3}[/tex]

From there, differentiating to get f''(x) is pretty straightforward.
 

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