Question Number 184757 by Shrinava last updated on 11/Jan/23 | ||
$$\mathrm{Which}\:\mathrm{function}\:\mathrm{has}\:\mathrm{a}\:\mathrm{crisis}\:\mathrm{point}? \\ $$$$\left.\mathrm{a}\right)\mathrm{y}=\mathrm{x}^{\mathrm{3}} +\mathrm{2x}+\mathrm{6} \\ $$$$\left.\mathrm{b}\right)\mathrm{y}=\sqrt[{\mathrm{4}}]{\mathrm{x}} \\ $$$$\left.\mathrm{c}\right)\mathrm{y}=\frac{\mathrm{15}}{\mathrm{x}} \\ $$$$\left.\mathrm{d}\right)\mathrm{y}=\mathrm{e}^{\boldsymbol{\mathrm{x}}} \\ $$$$\left.\mathrm{e}\right)\mathrm{y}=\sqrt[{\mathrm{3}}]{\mathrm{x}} \\ $$ | ||
Commented by mr W last updated on 11/Jan/23 | ||
$${i}\:{guess}\:{you}\:{mean}\:{critical}\:{point}. \\ $$ | ||
Commented by Shrinava last updated on 11/Jan/23 | ||
$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{professor} \\ $$ | ||
Answered by TheSupreme last updated on 11/Jan/23 | ||
$${b}\:{no} \\ $$$${c}\:{no} \\ $$$${d}\:{no} \\ $$$${e}\:{no} \\ $$$$ \\ $$$${a}...\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}=\mathrm{0}...\:{no}! \\ $$ | ||
Commented by Shrinava last updated on 11/Jan/23 | ||
$$\mathrm{Ser},\:\mathrm{so}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{right}\:\mathrm{answer}? \\ $$ | ||
Answered by pnuvu last updated on 11/Jan/23 | ||
$${i}\:{guess}\:{b}\:{and}\:{c} \\ $$ | ||
Commented by mr W last updated on 11/Jan/23 | ||
$${you}\:{lost}! \\ $$ | ||