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what is the stationary point of y=9x+6x^2-4x^3.need help asap

Sagot :

konrad509
[tex]y=9x+6x^2-4x^3\\ y'=9+12x-12x^2\\ 9+12x-12x^2=0\\ -12x^2+12x+9=0\\ 4x^2-4x-3=0\\ 4x^2-4x+1-4=0\\ (2x-1)^2=4\\ 2x-1=2 \vee 2x-1=-2\\ 2x=3 \vee 2x=-1\\ x=\frac{3}{2} \vee x=-\frac{1}{2}[/tex]
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