If S_{1}\; and\; S_{2} are respectively the sets of local minimum and local maximum points of the function,f(x)=9x^{4}+12x^{3}-36x^{2}+25,x\: \epsilon \: \mathbb{R}, then :
 

  • Option 1)

    S_{1}=\left \{ -2 \right \};S_{2}=\left \{ 0,1 \right \}

  • Option 2)

    S_{1}=\left \{ -2 ,1\right \};S_{2}=\left \{ 0 \right \}

  • Option 3)

    S_{1}=\left \{ -2 ,0\right \};S_{2}=\left \{ 1 \right \}

     

  • Option 4)

    S_{1}=\left \{ -1\right \};S_{2}=\left \{ 0,2 \right \}

 

Answers (1)

f(x)=9x^{4}+12x^{3}-36x^{2}+25\; \; \; \; \; x\: \epsilon \; \mathbb{R}

y=9x^{4}+12x^{3}-36x^{2}+25

\frac{dy}{dx}=36x^{3}+36x^{2}-72x

         =36(x^{3}+x^{2}-2)

         =36x(x^{3}+x-2)

          =36x(x+2)(x-1)

\frac{-\; \; \; +\; \; \; -\; \; \; +}{-2\; \; \; 0\; \; \; +1}

S_{1}=point\; of\; minima=\left \{ -2,1 \right \}

S_{2}=point\; of\; mixima=\left \{ 0 \right \}

 

 

 


Option 1)

S_{1}=\left \{ -2 \right \};S_{2}=\left \{ 0,1 \right \}

Option 2)

S_{1}=\left \{ -2 ,1\right \};S_{2}=\left \{ 0 \right \}

Option 3)

S_{1}=\left \{ -2 ,0\right \};S_{2}=\left \{ 1 \right \}

 

Option 4)

S_{1}=\left \{ -1\right \};S_{2}=\left \{ 0,2 \right \}

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