Question Number 75100 by Rio Michael last updated on 07/Dec/19
$$\:{find}\:{the}\:{intervals}\:{cor}\:{which}\:{the}\:{function} \\ $$$${h}\left({x}\right)\:=\:{x}^{\mathrm{3}} −\mathrm{3}{x}\:{is} \\ $$$$\left.{a}\right)\:{strickly}\:{increasing} \\ $$$$\left.{b}\right)\:{strickly}\:{decreasing} \\ $$$$ \\ $$
Answered by mr W last updated on 07/Dec/19
$${h}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}=\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\left({a}\right) \\ $$$${h}'\left({x}\right)=\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)>\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} >\mathrm{1} \\ $$$$\Rightarrow{x}<−\mathrm{1}\:{or}\:{x}>\mathrm{1} \\ $$$$\left({b}\right) \\ $$$${h}'\left({x}\right)=\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)<\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} <\mathrm{1} \\ $$$$\Rightarrow−\mathrm{1}<{x}<\mathrm{1} \\ $$
Answered by Kunal12588 last updated on 07/Dec/19
$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{3}{x} \\ $$$${y}={x}^{\mathrm{3}} −\mathrm{3}{x} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}=\mathrm{3}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right) \\ $$$${putting}\:\frac{{dy}}{{dx}}=\mathrm{0}\:{gives}\:\:{x}=\pm\mathrm{1} \\ $$$${which}\:{divides}\:{the}\:\:{number}\:{line}\:{into}\:\mathrm{3}\:{intervals} \\ $$$$\left(−\infty,−\mathrm{1}\right),\left(−\mathrm{1},\mathrm{1}\right)\:{and}\:\left(\mathrm{1},\infty\right) \\ $$$$\boldsymbol{{interval}}\:\:\:\:\mid\:\boldsymbol{{sign}}\:\boldsymbol{{of}}\:\:\boldsymbol{{f}}\:'\left(\boldsymbol{{x}}\right)\:\mid\:\boldsymbol{{nature}}\:\boldsymbol{{of}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right) \\ $$$$\left(−\infty,−\mathrm{1}\right)\:\:\mid\:\:\:\:\:\left(−\right)\left(−\right)\:\:\:\mid\:\:\:{strictly}\:{increasing} \\ $$$$\left(−\mathrm{1},\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\left(−\right)\left(+\right)\:\:\:\:\mid\:\:\:{strictly}\:{decreasing} \\ $$$$\left(\mathrm{1},\infty\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\left(+\right)\left(+\right)\:\:\:\:\:\mid\:\:\:{strictly}\:{increasing} \\ $$$$\therefore\:{f}\left({x}\right)\:{is}\:{strictly}\:{increasing}\:{in}\:\left(−\infty,−\mathrm{1}\right)\cup\left(\mathrm{1},\infty\right) \\ $$$${f}\left({x}\right)\:{is}\:{strictly}\:{decreasing}\:{in}\:\left(−\mathrm{1},\mathrm{1}\right) \\ $$
Commented by Rio Michael last updated on 07/Dec/19
$${wow}\:{sir}\:{i}\:{appreciate} \\ $$
Commented by peter frank last updated on 07/Dec/19
$${good}\: \\ $$