Question Number 159669 by zakirullah last updated on 19/Nov/21
$${find}\:{the}\:{relative}\:{maximum}\:{or}\:{minimum} \\ $$$${or}\:{neither}\:{at}\:{the}\:{given}\:{critical}\: \\ $$$${points}\:{of}\:{the}\:{function}? \\ $$$${f}^{'} \left({x}\right)=\mathrm{6}{x}\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{8}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{4}} ,\: \\ $$$${x}\:=\:\mathrm{1},\:{x}\:=\:\mathrm{2} \\ $$
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](https://www.tinkutara.com/question/Q159678.png)
$${Solution}: \\ $$$${f}^{'} \left({x}\right)=\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \left[\mathrm{3}+\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\right] \\ $$$${f}^{'} \left({x}\right)=\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${notice}\:{that}\:{the}\:\left({x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \:{is}\:\geqslant\mathrm{0} \\ $$$${so}\:{we}\:{only}\:{need}\:{to}\:{worry}\:{about}\:\mathrm{2}{x}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$−\:\:\:\:\underset{−\frac{\mathrm{1}}{\mathrm{2}}} {\bullet}\:\:\:\:+\:\:\:\:\:\underset{\mathrm{0}} {\bullet}\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\bullet}\:\:\:\:\:\:+ \\ $$$${x}=\mathrm{1}\:{and}\:{x}=\mathrm{2}\:{are}\:{after}\:\frac{\mathrm{1}}{\mathrm{2}}\:{so}\: \\ $$$${there}\:{is}\:{no}\:{local}\:{extrema} \\ $$$${at}\:{x}=\mathrm{1}\:{and}\:{x}=\mathrm{2} \\ $$$$ \\ $$
Commented by MohammadAzad last updated on 19/Nov/21
Commented by zakirullah last updated on 23/Nov/21
$${a}\:{boundle}\:{of}\:{thanks}\:{sir}\:{G} \\ $$