# point of inflection..blur..solution attached

• Mar 26th 2010, 10:39 AM
nikk
point of inflection..blur..solution attached
The equation as attach below. i have the problem on question (b). i confuse to determine the value of point of inflection because the i have the ans for
dy/dx = +ve the ans given is dy/dx = -ve. see the circle which i mark.

for your info, when i substitute x=-2 into dy/dx equation, i have the ans as 1/2 which is a +ve as a ans.

• Mar 26th 2010, 01:10 PM
HallsofIvy
dy/dx has four factors- $\displaystyle x^2$, $\displaystyle (x+ 1)^2$, $\displaystyle (x- 2)^2$, and $\displaystyle x^2- 2x- 6$. The first three are squared so are never negative- the sign of the derivative is the sign of $\displaystyle x^2- 2x- 6= x^2+ 2x+ 1- 7= (x+ 1)^2- 7$ which changes sign at $\displaystyle x= -1\pm\sqrt{7}$, the -1.65 and 3.65 you give. But that means that dy/dx> 0 for x> 3.65 or x< -1.65 and is negative for x between -1.65 and 3.65. It is certainly NOT true that the derivative is negative for all x< 0.
• Mar 26th 2010, 03:30 PM
nikk
Quote:

Originally Posted by HallsofIvy
dy/dx has four factors- $\displaystyle x^2$, $\displaystyle (x+ 1)^2$, $\displaystyle (x- 2)^2$, and $\displaystyle x^2- 2x- 6$. The first three are squared so are never negative- the sign of the derivative is the sign of $\displaystyle x^2- 2x- 6= x^2+ 2x+ 1- 7= (x+ 1)^2- 7$ which changes sign at $\displaystyle x= -1\pm\sqrt{7}$, the -1.65 and 3.65 you give. But that means that dy/dx> 0 for x> 3.65 or x< -1.65 and is negative for x between -1.65 and 3.65. It is certainly NOT true that the derivative is negative for all x< 0.

so it mean there is no inflection point here...???? because what i have CIRLE is wrong. so how should graph look like???
• Mar 29th 2010, 05:40 AM
nikk
Hope this is the ans. any comment ???