really need help on this
y= 1/3 x^3 + x^2-15x +3
y'= x^2 +2x-15 =0
(x+5) (x-3)
x=-5 x=3
y'' = 2x +2 =0
check for max or min
2(-5)+2 =-8< 0 local max
2(3) +2= 8 >o local min
check for poi
y'''= 2 not= 0 so we have a poi check if its stationary
2x+2=0 2X=-2 x=- 2/2 x=-1
non stationary point of infecton
so if the was x=0 we have a stationary point of inflection?
because i the book is they have this
y=(x+4)^3
y'=3(x+4)^2 =o x=-4
y''= 6(x+4)=0 x=-4
y'''=6 is not = to 0 we have a poi
the order of the derivative is an odd number so at x=-4 we have a poi since y=0 when x=-1 the point is stationary
bit confused because i thought that when the 2nd derviative result is calculated and x=0 then it is stationary when its x not = 0 nonstationary.
Hi matlondon,
no need to check the third derivative,
that's a different ball game.
If a point of inflexion is stationary, then the first and second derivatives are zero for the same x.
Calculate the first derivative.
This is zero at a maximum or minimum or saddle point.
The saddle point is another term for a stationary point of inflexion.
To know whether the point is a max, min or stationary point of inflexion,
use the value of x that causes the first derivative to be zero.
Plug this x into the 2nd derivative equation.
If the answer is negative, the point is a local maximum.
If the answer is positive, then it's a local minimum.
if the answer is zero, it's a stationary point of inflexion.
This applies to functions of a single variable, such as x.
It's more complex for functions of more than one variable.