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

In part (ii), we showed that the fixed points of ${\displaystyle g(x)={\frac {1}{2}}x(1-x)+x=-{\frac {1}{2}}x^{2}+{\frac {3}{2}}x}$ are ${\displaystyle x=0,1.}$ In order to classify these points, we first need to find the derivative of ${\displaystyle g(x)}$. Clearly,
${\displaystyle g'(x)=-x+{\frac {3}{2}}.}$
At the point ${\displaystyle x=0,}$ we have ${\displaystyle g'(0)={\frac {3}{2}}>1.}$ So, this fixed point is unstable.
At the point ${\displaystyle x=1,}$ we have ${\displaystyle g'(1)=-1+{\frac {3}{2}}={\frac {1}{2}}\in (0,1).}$ So, this fixed point is stable also.