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Question Number 32776 by NECx last updated on 02/Apr/18

$$Findtheoptimumpointsof$$$$thefunctiony=f\left(x\right)$$$$f\left(x\right)=2x^3-3x^2-36x+34$$

Answered by MJS last updated on 02/Apr/18

$$f^{\prime }\left(x\right)=0\wedge f^{″}\left(x\right)>0\Rightarrow min$$$$f^{\prime }\left(x\right)=0\wedge f^{″}\left(x\right)<0\Rightarrow max$$$$$$$$f^{\prime }\left(x\right)=6x^2-6x-36$$$$f^{″}\left(x\right)=12x-6$$$$$$$$f^{\prime }\left(x\right)=0$$$$x^2-x-6=0$$$$x=\frac{1}{2}\pm \sqrt{\frac{1}{4}+6}=\frac{1}{2}\pm \sqrt{\frac{25}{4}}=$$$$=\frac{1}{2}\pm \frac{5}{2}\Rightarrow x_1=-2;x_2=3$$$$f^{″}(-2)=-30\Rightarrow max=\left(\begin{array}{c}-2\\ 78\end{array}\right)$$$$f^{″}\left(3\right)=30\Rightarrow min=\left(\begin{array}{c}3\\ -47\end{array}\right)$$

Commented by NECx last updated on 02/Apr/18

$$thankyousomuch$$

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