View Full Version : Curve sketching: 2nd derivatives test
akotoh
Oct 1, 2009, 08:06 AM
Gud day! :)
I just want to know what is the correct answer in this problem:
Equation:
y = 1/3 x^3 - 1/2 x^2 - 2x
the point of inflection that I got is (1/2 , -3/4) but according to the book its (1/2, -13/12). How do they get that?
thanks! :)
ArcSine
Oct 1, 2009, 08:14 AM
Just evaluate the original equation for x = 1/2.
radiation
Oct 4, 2009, 03:17 AM
I think thr is some error in your ans.. for x=1/2, the only value for y is -13/12..
galactus
Oct 4, 2009, 04:08 AM
Gud day! :)
I just want to know what is the correct answer in this problem:
Equation:
y = 1/3 x^{3} - 1/2 x^{2} - 2x
the point of inflection that i got is (1/2 , -3/4) but according to the book its (1/2, -13/12). How do they get that?
thanks! :)
They found the second derivative, set to 0 and solved for x.
y'=x^{2}-x-2
y''=2x-1
2x-1=0\Rightarrow x=\frac{1}{2}
Plug back into original equation:
\frac{1}{3}(\frac{1}{2})^{3}-\frac{1}{2}(\frac{1}{2})^{2}-2(\frac{1}{2})=\frac{-13}{12}