Question

Ex 2.6
11. Find any stationary points on the curve y=(x-3) √(4-x) and determine their nature

Answer 1

`y=(x-3) √(4-x)\\ x\leq4\\\\ y'=√(4-x)+(x-3)\cdot(1)/(2√(4-x))\cdot(-1)\\ y'=√(4-x)+(-x+3)/(2√(4-x))\\ y'=(1)/(2√(4-x))(2(4-x)-x+3)\\ y'=(1)/(2√(4-x))(8-2x-x+3)\\ y'=(1)/(2√(4-x))(-3x+11)\\ y'=-(1)/(2√(4-x))(3x-11)`


`-(1)/(2√(4-x))(3x-11)=0\\ 3x-11=0\\ 3x=11\\ x=(11)/(3)\leftarrow \text{ stationary point}\\\\ \forall x\in(-\infty,(11)/(3))\ y'>0\Rightarrow y\\earrow\\ \forall x\in((11)/(3),\infty)\ y'<0\Rightarrow y\searrow\\\Downarrow\\ y\left((11)/(3)\right)=y_(max)`