# Thread: Nature of a point

1. ## Nature of a point

Given that $\displaystyle \frac{dy}{dx}=6(x-1)(x-2)$ and the coordinates of the stationery points are $\displaystyle (1,1)$ and $\displaystyle (2,0)$, determine the nature of these stationery points.

2. Originally Posted by Punch
Given that $\displaystyle \frac{dy}{dx}=6(x-1)(x-2)$ and the coordinates of the stationery points are $\displaystyle (1,1)$ and $\displaystyle (2,0)$, determine the nature of these stationery points.
The x-coordinates of these points will make the drerivative = zero.
so what do you think about thier nature?

3. Originally Posted by General
The x-coordinates of these points will make the drerivative = zero.
so what do you think about thier nature?
The answer is either minimum point, maximum point or a point of inflexion.
The answer can be derived by substituting in values which are < and > the x-coordinate of the stationery point to determine if the stationery point is a maximum, minimum or a point of inflexion.

However, the answer I have gotten does not tally with the answer from the book.

4. Originally Posted by Punch
The answer is either minimum point, maximum point or a point of inflexion.
The answer can be derived by substituting in values which are < and > the x-coordinate of the stationery point to determine if the stationery point is a maximum, minimum or a point of inflexion.

However, the answer I have gotten does not tally with the answer from the book.
You did not specify the choices.
find $\displaystyle \frac{d^2y}{dx^2}$ and try to see if one of two points is a point of inflection or not.
Also, how can you know if the point has maximum/minimum value for its function?

5. okay, here is an example...
If by substituting values of x < x-coordinate into dy/dx, you get a negative gradient, this means that it is down sloping.
If by substituting values of x > x-coordinate into dy/dx, you get a positive gradient, this means that it is up sloping.

So we can conclude that the x-coordinate point is a minimum point as the gradient is negative before increasing after the stationery point where dy/dx=0.

6. Yes, now here is your function: dy/dx= 6(x-1)(x-2).

If x< 1, both x- 1 and x- 2 are negative. What does that tell you about their product?

If 1< x< 2, x- 1 is positive while x- 2 is negative. What does that tell you about their product?

If 2< x, both x- 1 and x- 2 are positive. What does that tell you about their product?