Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 159669 by zakirullah last updated on 19/Nov/21

find the relative maximum or minimum  or neither at the given critical   points of the function?  f^′ (x)=6x(x^2 −4)^4 (x^2 −1)^2 +8x(x^2 −1)^3 (x^2 −4)^4 ,   x = 1, x = 2

$${find}\:{the}\:{relative}\:{maximum}\:{or}\:{minimum} \\ $$$${or}\:{neither}\:{at}\:{the}\:{given}\:{critical}\: \\ $$$${points}\:{of}\:{the}\:{function}? \\ $$$${f}^{'} \left({x}\right)=\mathrm{6}{x}\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{8}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{4}} ,\: \\ $$$${x}\:=\:\mathrm{1},\:{x}\:=\:\mathrm{2} \\ $$

Answered by MohammadAzad last updated on 19/Nov/21

Solution:  f^′ (x)=2x(x^2 −4)^4 (x^2 −1)^2 [3+4(x^2 −1)]  f^′ (x)=2x(x^2 −4)^4 (x^2 −1)^2 (4x^2 −1)  notice that the (x^2 −4)^4 (x^2 −1)^2  is ≥0  so we only need to worry about 2x(4x^2 −1)  −    •_(−(1/2))     +     •_0         −      •_(1/2)       +  x=1 and x=2 are after (1/2) so   there is no local extrema  at x=1 and x=2

$${Solution}: \\ $$$${f}^{'} \left({x}\right)=\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \left[\mathrm{3}+\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\right] \\ $$$${f}^{'} \left({x}\right)=\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${notice}\:{that}\:{the}\:\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \:{is}\:\geqslant\mathrm{0} \\ $$$${so}\:{we}\:{only}\:{need}\:{to}\:{worry}\:{about}\:\mathrm{2}{x}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$−\:\:\:\:\underset{−\frac{\mathrm{1}}{\mathrm{2}}} {\bullet}\:\:\:\:+\:\:\:\:\:\underset{\mathrm{0}} {\bullet}\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\bullet}\:\:\:\:\:\:+ \\ $$$${x}=\mathrm{1}\:{and}\:{x}=\mathrm{2}\:{are}\:{after}\:\frac{\mathrm{1}}{\mathrm{2}}\:{so}\: \\ $$$${there}\:{is}\:{no}\:{local}\:{extrema} \\ $$$${at}\:{x}=\mathrm{1}\:{and}\:{x}=\mathrm{2} \\ $$$$ \\ $$

Commented by MohammadAzad last updated on 19/Nov/21

Commented by zakirullah last updated on 23/Nov/21

a boundle of thanks sir G

$${a}\:{boundle}\:{of}\:{thanks}\:{sir}\:{G} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com