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.

Printable View

- Mar 3rd 2010, 02:57 AMPunchNature 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.

- Mar 3rd 2010, 03:05 AMGeneral
- Mar 3rd 2010, 03:19 AMPunch
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. - Mar 3rd 2010, 03:23 AMGeneral
- Mar 3rd 2010, 03:27 AMPunch
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. - Mar 3rd 2010, 05:47 AMHallsofIvy
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?