Question Number 123812 by benjo_mathlover last updated on 28/Nov/20
$${Within}\:{the}\:{interval}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi\:,{find}\: \\ $$$${the}\:{critical}\:{points}\:{of}\: \\ $$$${f}\left({x}\right)=\:\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{sin}\:{x}−\mathrm{1}\:.\:{Identify} \\ $$$${the}\:{open}\:{interval}\:{on}\:{which}\:{f}\:{is}\: \\ $$$${increasing}\:{and}\:{decreasing}\:.\:{Find} \\ $$$${the}\:{function}'{s}\:{local}\:{and}\:{absolute} \\ $$$${extreme}\:{values}. \\ $$
Answered by liberty last updated on 28/Nov/20
![The function f is continous over [ 0,2π ] and differentiable over (0,2π) so the critical points occur at the zeros of f ′ in (0,2π) . We find f ′(x)= (2sin x−1)cos x . The first derivative is zero if only if sin x=(1/2) or cos x=0 . So the critical points of f in (0,2π) are x=(π/6); ((5π)/6); (π/2); ((3π)/2) they partition [0,2π] into open intervals as follows interval : (0,(π/6)) ((π/6),(π/2)) ((π/2),((5π)/6)) (((5π)/6),((3π)/2)) (((3π)/2),2π) sign of f ′ : − + − + − behavior of f : dec inc dec inc dec there is a local minimum value of f((π/6))=−(5/4), a local maximum value of f((π/2))=−1, another local minimum value of f(((5π)/6))=−(5/4) and another local maximum value of f(((3π)/2))= 1. The endpoint value are f(0)=f(2π)=−1. The absolute minimum in [ 0,2π ] is −(5/4) occuring at x=(π/6) and x=((5π)/6); the absolute maximum is 1 occuring at x=((3π)/2).](Q123818.png)
$${The}\:{function}\:{f}\:{is}\:{continous}\:{over}\:\left[\:\mathrm{0},\mathrm{2}\pi\:\right]\:{and}\:{differentiable}\:{over}\:\left(\mathrm{0},\mathrm{2}\pi\right)\: \\ $$$${so}\:{the}\:{critical}\:{points}\:{occur}\:{at}\:{the}\:{zeros}\:{of}\:{f}\:'\:{in}\:\left(\mathrm{0},\mathrm{2}\pi\right)\:.\:{We}\:{find}\: \\ $$$${f}\:'\left({x}\right)=\:\left(\mathrm{2sin}\:{x}−\mathrm{1}\right)\mathrm{cos}\:{x}\:.\:{The}\:{first}\:{derivative}\:{is}\:{zero}\:{if}\:{only}\:{if}\:\mathrm{sin}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${or}\:\mathrm{cos}\:{x}=\mathrm{0}\:.\:{So}\:{the}\:{critical}\:{points}\:{of}\:{f}\:{in}\:\left(\mathrm{0},\mathrm{2}\pi\right)\:{are}\:{x}=\frac{\pi}{\mathrm{6}};\:\frac{\mathrm{5}\pi}{\mathrm{6}};\:\frac{\pi}{\mathrm{2}};\:\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${they}\:{partition}\:\left[\mathrm{0},\mathrm{2}\pi\right]\:{into}\:{open}\:{intervals}\:{as}\:{follows}\: \\ $$$${interval}\::\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{0},\frac{\pi}{\mathrm{6}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\pi}{\mathrm{6}},\frac{\pi}{\mathrm{2}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\pi}{\mathrm{2}},\frac{\mathrm{5}\pi}{\mathrm{6}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{5}\pi}{\mathrm{6}},\frac{\mathrm{3}\pi}{\mathrm{2}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}},\mathrm{2}\pi\right) \\ $$$${sign}\:{of}\:{f}\:'\::\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:− \\ $$$${behavior}\:{of}\:{f}\::\:\:\:\:\:\:\:\:\:\:\:\:\:{dec}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{inc}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{dec}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{inc}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{dec} \\ $$$${there}\:{is}\:{a}\:{local}\:{minimum}\:{value}\:{of}\:{f}\left(\frac{\pi}{\mathrm{6}}\right)=−\frac{\mathrm{5}}{\mathrm{4}},\:{a}\:{local}\:{maximum}\:{value}\:{of} \\ $$$${f}\left(\frac{\pi}{\mathrm{2}}\right)=−\mathrm{1},\:{another}\:{local}\:{minimum}\:{value}\:{of}\:{f}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)=−\frac{\mathrm{5}}{\mathrm{4}}\:{and}\:{another}\:{local} \\ $$$${maximum}\:{value}\:{of}\:{f}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)=\:\mathrm{1}.\:{The}\:{endpoint}\:{value}\:{are}\:{f}\left(\mathrm{0}\right)={f}\left(\mathrm{2}\pi\right)=−\mathrm{1}.\:{The} \\ $$$${absolute}\:{minimum}\:{in}\:\left[\:\mathrm{0},\mathrm{2}\pi\:\right]\:{is}\:−\frac{\mathrm{5}}{\mathrm{4}}\:{occuring}\:{at}\:{x}=\frac{\pi}{\mathrm{6}}\:{and}\:{x}=\frac{\mathrm{5}\pi}{\mathrm{6}};\:{the}\:{absolute} \\ $$$${maximum}\:{is}\:\mathrm{1}\:{occuring}\:{at}\:{x}=\frac{\mathrm{3}\pi}{\mathrm{2}}. \\ $$