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Question Number 197822 by mokys last updated on 30/Sep/23

$$findmaximumof\mid z^2+2z-3\mid ?$$

Commented by mokys last updated on 30/Sep/23

$$theansweris\frac{8}{\sqrt{3}}$$

Commented by mokys last updated on 30/Sep/23

$$yes$$

Answered by a.lgnaoui last updated on 30/Sep/23

$$\{\begin{array}{l}\mathrm{f}\left(\mathrm{z}\right)={\mathrm{z}}^2+2\mathrm{z}-3\mathrm{z}\in ]-\mathrm{\infty },-3][1+\mathrm{\infty }\\ \mathrm{f}\left(\mathrm{z}\right)=3-2\mathrm{z}-{\mathrm{z}}^2\mathrm{z}\in [-3,1]\end{array}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{z}\right)=0\mathrm{z}=-1\in [-3,1]\Rightarrow \mathrm{f}(-1)=4$$$$\mathrm{so}\mathrm{the}\mathrm{maximum}\mathrm{relative}\mathrm{is}4$$

Commented by a.lgnaoui last updated on 30/Sep/23

$$\mathrm{for}\mathrm{f}\left(\mathrm{z}\right)={\mathrm{z}}^2+2\mathrm{z}-3(\mathrm{z}\in ]-\mathrm{\infty },-3]\bigsqcup [1,+\mathrm{\infty }\left[\right)$$$$\mathrm{the}\mathrm{maximum}\mathrm{absolu}\mathrm{is}+\mathrm{\infty }$$

Commented by a.lgnaoui last updated on 30/Sep/23

Commented by mahdipoor last updated on 30/Sep/23

$$f(-3)=f\left(1\right)=0isabsoluteminimum$$$$f(-1)=4islocalmaximum$$$$f(\pm \mathrm{\infty })=+\mathrm{\infty }isabsolutemaximum$$

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