1. ## question about points of inflection derivatives (urgent)

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.

2. really need help on this

3. Originally Posted by matlondon
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,

$\displaystyle f(x)=(x+4)^3$

$\displaystyle f'(x)=3(x+4)^2=0\ for\ x=-4$

$\displaystyle f''(x)=6(x+4)=0\ for\ x=-4$

$\displaystyle f''(x)$ is neither positive or negative at x=-4, so it is a saddle point, a stationary point of inflexion.

That's as far as you need to go to know the difference.

$\displaystyle f(x)=\frac{x^3}{3}+x^2-15x+3$

$\displaystyle f'(x)=x^2+2x-15=(x+5)(x-3)$

This is zero at $\displaystyle x=-5,\ x=3$

$\displaystyle f''(x)=2x+2=max\ at\ x=-5,\ min\ at\ x=3$

$\displaystyle f''(x)=0\ for\ x=-1$

There is a non-stationary point of inflexion at $\displaystyle x=-1$

4. f ' ' ' = 2 not equal to 0 so we have a poi
not we plug in the x-1 into the first derivative to see if it stationary or non.

f'(x)=-1^2+2(-1)-15= -18 this is non stationary

so if the same the equation and the answer was zero then it would a a stationary poi?

5. ## stationary point

Stationary points only when the first derivative =0

6. Originally Posted by matlondon

f ' ' ' = 2 not equal to 0 so we have a poi
not we plug in the x-1 into the first derivative to see if it stationary or non.

f'(x)=-1^2+2(-1)-15= -18 this is non stationary

so if the same the equation and the answer was zero then it would a a stationary poi?
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.