# Thread: Maxima/minima/points of inflection

1. ## Maxima/minima/points of inflection

Hi all,
Can somebody tell me if my solution is correct?

y= (x^3)-[(3/2)*(x^2)]-6x+9

Find the stationary values, determine their nature, and find the point of inflection.

Max= (-1,12.5)
Min= (2, -1)
Point of inflection= (1/2, 5.75)

thanks
John

Solution:

At stationary points dy/dx (slope=0)

dy/dx=0=3x^2 -3x-6

solve quadratic to find x values

X=2, -1

Substitute into expression for y gives corresponding y values of (-1,12.5)

The second derivative=6x-3

The second derivative for x=2 =9 which is greater than 0 so this is the minimum point
The second derivative for x=-1 =-9 which is less than 0 so this is the maximum point

At the point of inflection the second derivative=0

d2y/dx2=0=6x-3 therefore x=1/2

Substituting x=1/2 into the expression for y gives y value of 5.75.

Therefore the point of inflection= (1/2, 5.75)

2. ## Re: Maxima/minima/points of inflection

Looks fine! To determine if you're dealing with a minimum or maximum you can also make a sign table.

thanks Siron