Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 147980 by mathdanisur last updated on 24/Jul/21

f(x) = 4x^4  - 2x^2  + 17  Find the maximum point of the  function

$${f}\left({x}\right)\:=\:\mathrm{4}{x}^{\mathrm{4}} \:-\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{17} \\ $$$${Find}\:{the}\:{maximum}\:{point}\:{of}\:{the} \\ $$$${function} \\ $$

Answered by iloveisrael last updated on 25/Jul/21

f ′(x)=16x^3 −4x = 0  ⇒4x(4x^2 −1)=0  ⇒x=0 , x = ± (1/2)  f ′′(x)=48x^2 −4 < 0 for x = 0  local max when x = 0 , local max value = 17  local max point at (0,17)

$$\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{16x}^{\mathrm{3}} −\mathrm{4x}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{4x}\left(\mathrm{4x}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{0}\:,\:\mathrm{x}\:=\:\pm\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{f}\:''\left(\mathrm{x}\right)=\mathrm{48x}^{\mathrm{2}} −\mathrm{4}\:<\:\mathrm{0}\:\mathrm{for}\:\mathrm{x}\:=\:\mathrm{0} \\ $$$$\mathrm{local}\:\mathrm{max}\:\mathrm{when}\:\mathrm{x}\:=\:\mathrm{0}\:,\:\mathrm{local}\:\mathrm{max}\:\mathrm{value}\:=\:\mathrm{17} \\ $$$$\mathrm{local}\:\mathrm{max}\:\mathrm{point}\:\mathrm{at}\:\left(\mathrm{0},\mathrm{17}\right)\: \\ $$$$ \\ $$

Commented by mathdanisur last updated on 25/Jul/21

Thanks Ser

$${Thanks}\:{Ser} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com