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Question Number 69500 by TawaTawa last updated on 24/Sep/19

Answered by MJS last updated on 24/Sep/19

$$\mathrm{the}\mathrm{shape}\mathrm{of}\mathrm{the}\mathrm{graph}\mathrm{looks}\mathrm{like}\mathrm{a}\mathrm{modified}$$$$f\left(x\right)=x^4-x^2$$$$f\left(0\right)\approx -7\Rightarrow f\left(x\right)\approx x^4-x^2-7$$$$\mathrm{the}\mathrm{zeros}\mathrm{are}\mathrm{at}\approx \pm 1.2\Rightarrow$$$$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{put}x=at$$$$f\left(t\right)=a^4{t}^4-a^2{t}^2-7$$$$t=1.2\Rightarrow 2.0736a^4-1.44a^2-7=0\Rightarrow a\approx 1.48898$$$$\mathrm{let}a=\frac{3}{2}$$$$f\left(x\right)\approx \frac{81}{16}{x}^4-\frac{9}{4}{x}^2-7$$

Commented by TawaTawa last updated on 24/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by MJS last updated on 24/Sep/19

$$y=\frac{x-1}{{(x+1)}^2}$$$$y^{\prime }=-\frac{x-3}{{(x+1)}^3}=0\Rightarrow x=3$$$$y^{″}=2\frac{x-5}{{(x+1)}^4}\mathrm{with}x=3\Rightarrow y^{″}<0\Rightarrow \max \mathrm{at}\left(\begin{array}{c}3\\ \frac{1}{8}\end{array}\right)$$$$\mathrm{the}\mathrm{range}\mathrm{is}-\mathrm{\infty }<x⩽\frac{1}{8}$$

Commented by TawaTawa last updated on 24/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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