Ask Me Help Desk - Curve sketching: 2nd derivatives test

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! :)

Just evaluate the original equation for x = 1/2.

I think thr is some error in your ans.. for x=1/2, the only value for y is -13/12..

Quote:

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:

$\frac{1}{3}(\frac{1}{2})^{3}-\frac{1}{2}(\frac{1}{2})^{2}-2(\frac{1}{2})=\frac{-13}{12}$