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

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