Question Number 197822 by mokys last updated on 30/Sep/23
$${find}\:{maximum}\:{of}\:\mid{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{3}\mid\:? \\ $$
Commented by mokys last updated on 30/Sep/23
$${the}\:{answer}\:{is}\:\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}} \\ $$
Commented by mokys last updated on 30/Sep/23
$${yes} \\ $$
Answered by a.lgnaoui last updated on 30/Sep/23
![{ ((f(z)=z^2 +2z−3 z∈]−∞,−3] [1+ ∞)),((f(z)=3−2z−z^2 z∈[−3,1])) :} f^′ (z)=0 z=−1∈[−3,1]⇒ f(−1)=4 so the maximum relative is 4](Q197828.png)
$$\begin{cases}{\left.\mathrm{f}\left.\left(\mathrm{z}\right)=\mathrm{z}^{\mathrm{2}} +\mathrm{2z}−\mathrm{3}\:\:\:\:\:\mathrm{z}\in\right]−\infty,−\mathrm{3}\right]\:\left[\mathrm{1}+\:\:\infty\right.}\\{\mathrm{f}\left(\mathrm{z}\right)=\mathrm{3}−\mathrm{2z}−\mathrm{z}^{\mathrm{2}} \:\:\:\:\:\mathrm{z}\in\left[−\mathrm{3},\mathrm{1}\right]}\end{cases} \\ $$$$\mathrm{f}^{'} \left(\mathrm{z}\right)=\mathrm{0}\:\:\:\mathrm{z}=−\mathrm{1}\in\left[−\mathrm{3},\mathrm{1}\right]\Rightarrow\:\:\:\mathrm{f}\left(−\mathrm{1}\right)=\mathrm{4} \\ $$$$\mathrm{so}\:\:\mathrm{the}\:\mathrm{maximum}\:\:\mathrm{relative}\:\mathrm{is}\:\mathrm{4} \\ $$
Commented by a.lgnaoui last updated on 30/Sep/23
![for f(z)=z^2 +2z−3(z∈]−∞,−3]⊔[1,+∞[) the maximum absolu is +∞](Q197829.png)
$$\left.\mathrm{for}\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{z}^{\mathrm{2}} +\mathrm{2z}−\mathrm{3}\left(\mathrm{z}\in\right]−\infty,−\mathrm{3}\right]\sqcup\left[\mathrm{1},+\infty\left[\right)\right. \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\:\mathrm{absolu}\:\mathrm{is}\:+\infty \\ $$
Commented by a.lgnaoui last updated on 30/Sep/23
Commented by mahdipoor last updated on 30/Sep/23
$${f}\left(−\mathrm{3}\right)={f}\left(\mathrm{1}\right)=\mathrm{0}\:\:\:{is}\:\:{absolute}\:{minimum} \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{4}\:{is}\:{local}\:{maximum} \\ $$$${f}\left(\pm\infty\right)=+\infty\:\:{is}\:{absolute}\:{maximum} \\ $$