Question Number 75368 by vishalbhardwaj last updated on 10/Dec/19
![Find the interval for which the function f(x) = sinx + cosx, for x∈ [0, 2π] is strictly inceasing and srictly decreasing ?](https://www.tinkutara.com/question/Q75368.png)
$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{interval}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{function} \\ $$$$\:{f}\left({x}\right)\:=\:{sinx}\:+\:{cosx},\:\mathrm{for}\:\:{x}\in\:\left[\mathrm{0},\:\mathrm{2}\pi\right] \\ $$$$\mathrm{is}\:\mathrm{strictly}\:\mathrm{inceasing}\:\mathrm{and}\:\mathrm{srictly}\:\mathrm{decreasing}\:? \\ $$
Commented by vishalbhardwaj last updated on 10/Dec/19
$$\mathrm{please}\:\mathrm{sir},\:\mathrm{write}\:\mathrm{complete}\:\mathrm{explanation} \\ $$$$\mathrm{of}\:\mathrm{intervals} \\ $$
Answered by MJS last updated on 10/Dec/19
; ((5π)/4)[ f(x) strictly decreasing ]((5π)/4); 2π] f(x) strictly increasing](https://www.tinkutara.com/question/Q75371.png)
$${f}'\left({x}\right)=\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x} \\ $$$${f}'\left({x}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:{x}\:=\mathrm{cos}\:{x} \\ $$$${x}=\frac{\pi}{\mathrm{4}}\vee{x}=\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$$${f}''\left({x}\right)=−\mathrm{sin}\:{x}\:−\mathrm{cos}\:{x} \\ $$$${f}''\left(\frac{\pi}{\mathrm{4}}\right)<\mathrm{0}\:\Rightarrow\:\mathrm{maximum} \\ $$$${f}''\left(\frac{\mathrm{5}\pi}{\mathrm{4}}\right)>\mathrm{0}\:\Rightarrow\:\mathrm{minimum} \\ $$$$\left[\mathrm{0};\:\frac{\pi}{\mathrm{4}}\left[\:{f}\left({x}\right)\:\mathrm{strictly}\:\mathrm{increasing}\right.\right. \\ $$$$\left.\right]\frac{\pi}{\mathrm{4}};\:\frac{\mathrm{5}\pi}{\mathrm{4}}\left[\:{f}\left({x}\right)\:\mathrm{strictly}\:\mathrm{decreasing}\right. \\ $$$$\left.\right]\left.\frac{\mathrm{5}\pi}{\mathrm{4}};\:\mathrm{2}\pi\right]\:{f}\left({x}\right)\:\mathrm{strictly}\:\mathrm{increasing} \\ $$
Commented by vishalbhardwaj last updated on 10/Dec/19
$$\mathrm{sir}\:\mathrm{please}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{first}\:\mathrm{derivative}\:\mathrm{rule} \\ $$$$\mathrm{and}\:\mathrm{please}\:\mathrm{explane}\:\mathrm{by}\:\mathrm{taking}\:\mathrm{any}\:\mathrm{value}\:\mathrm{in}\: \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{intervals} \\ $$
Commented by MJS last updated on 10/Dec/19
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{this}\:\mathrm{rule}.\:\mathrm{the}\:\mathrm{proper}\:\mathrm{way}\:\mathrm{I}\:\mathrm{used} \\ $$$$\mathrm{is}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{zeros}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{derivate}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{extremes}\:\mathrm{and}\:\mathrm{then}\:\mathrm{check}\:\mathrm{the}\:\mathrm{second}\:\mathrm{derivate} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{if}\:\mathrm{these}\:\mathrm{are}\:\mathrm{maxima}\:\mathrm{or}\:\mathrm{minima} \\ $$$${f}'\left({p}\right)=\mathrm{0}\wedge{f}''\left({p}\right)<\mathrm{0}\:\Leftrightarrow\:\mathrm{max}\:\mathrm{at}\:{p} \\ $$$${f}'\left({p}\right)=\mathrm{0}\wedge{f}''\left({p}\right)>\mathrm{0}\:\Leftrightarrow\:\mathrm{min}\:\mathrm{at}\:{p} \\ $$
Answered by Kunal12588 last updated on 10/Dec/19
![f ′(x)=cos x − sin x=(√2)cos((π/4)+x) f ′(x)=0 ⇒cos((π/4)+x)=0 ⇒x=2nπ±(π/2)−(π/4)=2nπ+(π/4),2nπ−((3π)/4) which divides [0,2π] into 3 intervals [0,(π/4)),((π/4),((5π)/4)),(((5π)/4),2π] interval ∣sign of f ′(x)∣nature of f(x) [0,(π/4)) ∣ (+) ∣ strictly increasing ((π/4),((5π)/4)) ∣ (−) ∣ strictly decreasing (((5π)/4),2π] ∣ (+) ∣ stictly increasing ∴ f(x) is increasing in [0,(π/4))∪(((5π)/4),2π] f(x) is decreasing in ((π/4),((5π)/4)) [Note: this answer is not valid for every value of x, only for the domain [0,2π] ]](https://www.tinkutara.com/question/Q75413.png)
$${f}\:'\left({x}\right)={cos}\:{x}\:−\:{sin}\:{x}=\sqrt{\mathrm{2}}{cos}\left(\frac{\pi}{\mathrm{4}}+{x}\right) \\ $$$${f}\:'\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow{cos}\left(\frac{\pi}{\mathrm{4}}+{x}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{4}},\mathrm{2}{n}\pi−\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$${which}\:{divides}\:\left[\mathrm{0},\mathrm{2}\pi\right]\:{into}\:\mathrm{3}\:{intervals} \\ $$$$\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right),\left(\frac{\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}}\right),\left(\frac{\mathrm{5}\pi}{\mathrm{4}},\mathrm{2}\pi\right] \\ $$$$\boldsymbol{{interval}}\:\mid\boldsymbol{{sign}}\:\boldsymbol{{of}}\:\boldsymbol{{f}}\:'\left(\boldsymbol{{x}}\right)\mid\boldsymbol{{nature}}\:\boldsymbol{{of}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right) \\ $$$$\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right)\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\left(+\right)\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{strictly}\:{increasing} \\ $$$$\left(\frac{\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}}\right)\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\left(−\right)\:\:\:\:\:\:\:\:\:\:\:\mid\:{strictly}\:{decreasing} \\ $$$$\left(\frac{\mathrm{5}\pi}{\mathrm{4}},\mathrm{2}\pi\right]\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\left(+\right)\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:{stictly}\:{increasing} \\ $$$$\therefore\:{f}\left({x}\right)\:{is}\:{increasing}\:{in}\:\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right)\cup\left(\frac{\mathrm{5}\pi}{\mathrm{4}},\mathrm{2}\pi\right]\: \\ $$$${f}\left({x}\right)\:{is}\:{decreasing}\:{in}\:\left(\frac{\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}}\right) \\ $$$$\left[\boldsymbol{{Note}}:\:{this}\:{answer}\:{is}\:{not}\:{valid}\:{for}\:{every}\right. \\ $$$$\left.{value}\:{of}\:{x},\:{only}\:{for}\:{the}\:{domain}\:\left[\mathrm{0},\mathrm{2}\pi\right]\:\:\right] \\ $$