Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 194088 by cortano12 last updated on 27/Jun/23

Answered by horsebrand11 last updated on 27/Jun/23

$${\mathrm{y}}^2=(2+(\mathrm{x}+{\mathrm{x}}^2))(1-(\mathrm{x}+{\mathrm{x}}^2))$$$$\mathrm{let}\mathrm{x}+{\mathrm{x}}^2=\mathrm{u}$$$${\mathrm{y}}^2=(2+\mathrm{u})(1-\mathrm{u})$$$${\mathrm{y}}^2=-{\mathrm{u}}^2-\mathrm{u}+2$$$${2\mathrm{yy}}^{\prime }=(-2\mathrm{u}-1){\mathrm{u}}^{\prime }$$$${\mathrm{y}}^{\prime }=\frac{(-2\mathrm{x}-{2\mathrm{x}}^2-1)(1+2\mathrm{x})}{2\sqrt{(2+\mathrm{x}+{\mathrm{x}}^2)(1-\mathrm{x}-{\mathrm{x}}^2)}}=0$$$$\iff ({2\mathrm{x}}^2+2\mathrm{x}+1)(1+2\mathrm{x})=0$$$$\mathrm{x}=-\frac{1}{2}\Rightarrow {\mathrm{f}}_{\max }=\frac{\sqrt{35}}{4}$$$${\mathrm{f}}_{\min }=0\mathrm{when}1-\mathrm{x}-{\mathrm{x}}^2=0$$$$\Rightarrow {\mathrm{x}}^2+\mathrm{x}-1=0$$$$\Rightarrow {(\mathrm{x}+\frac{1}{2})}^2=\frac{5}{4}$$$$\Rightarrow \mathrm{x}=\frac{\pm \sqrt{5}-1}{2}$$

Commented by mnjuly1970 last updated on 27/Jun/23

$$merceysir$$

Answered by Frix last updated on 27/Jun/23

$$0⩽f\left(x\right)⩽\frac{\sqrt{35}}{4}$$

Answered by Subhi last updated on 27/Jun/23

$$(1-x-x^2)\Rrightarrow maxvalueatf\left(\frac{-b}{2a}\right)=f\left(\frac{-1}{2}\right)=\frac{5}{4}$$$$\sqrt{(1-x-x^2)(2+x+x^2)}\Rrightarrow 1-x-x^2⩾0$$$$\therefore theminivalueis0$$$$x^2+x-1⩽0$$$$(x-\frac{-1+\sqrt{5}}{2})(x+\frac{1+\sqrt{5}}{2})⩽0$$$$x\in [\frac{-1-\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2}]$$$$f_{max}isatx=\frac{-1}{2},givesthemaxvalueof(1-x-x^2)$$$$f_{max}=\sqrt{\frac{5}{4}.\frac{7}{4}}=\frac{\sqrt{35}}{4}$$$$f_{max}+f_{mini}=\frac{\sqrt{35}}{4}$$

Commented by mnjuly1970 last updated on 27/Jun/23

$$perfectsir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com