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

Sagot :

konrad509
[tex] y=(x-3) \sqrt{4-x}\\ x\leq4\\\\ y'=\sqrt{4-x}+(x-3)\cdot\dfrac{1}{2\sqrt{4-x}}\cdot(-1)\\ y'=\sqrt{4-x}+\dfrac{-x+3}{2\sqrt{4-x}}\\ y'=\dfrac{1}{2\sqrt{4-x}}(2(4-x)-x+3)\\ y'=\dfrac{1}{2\sqrt{4-x}}(8-2x-x+3)\\ y'=\dfrac{1}{2\sqrt{4-x}}(-3x+11)\\ y'=-\dfrac{1}{2\sqrt{4-x}}(3x-11)[/tex]

[tex]-\dfrac{1}{2\sqrt{4-x}}(3x-11)=0\\ 3x-11=0\\ 3x=11\\ x=\dfrac{11}{3}\leftarrow \text{ stationary point}\\\\ \forall x\in(-\infty,\frac{11}{3})\ y'>0\Rightarrow y\nearrow\\ \forall x\in(\frac{11}{3},\infty)\ y'<0\Rightarrow y\searrow\\\Downarrow\\ y\left(\dfrac{11}{3}\right)=y_{max} [/tex]
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