Science:Math Exam Resources/Courses/MATH103/April 2016/Question 06 (b) (iii)/Solution 1

In part (ii), we showed that the fixed points of $g(x)={\frac {1}{2}}x(1-x)+x=-{\frac {1}{2}}x^{2}+{\frac {3}{2}}x$ are $x=0,1.$ In order to classify these points, we first need to find the derivative of $g(x)$. Clearly,

$g'(x)=-x+{\frac {3}{2}}.$

At the point $x=0,$ we have $g'(0)={\frac {3}{2}}>1.$ So, this fixed point is unstable.

At the point $x=1,$ we have $g'(1)=-1+{\frac {3}{2}}={\frac {1}{2}}\in (0,1).$ So, this fixed point is stable also.