Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 12601 by @ANTARES_VY last updated on 26/Apr/17

$$\mathbf{y}=\frac{{\mathbf{x}}^3}{3}+2{\mathbf{x}}^2-5\mathbf{x}+7\mathbf{find}\mathbf{the}$$$$\mathbf{critical}\mathbf{points}\mathbf{of}\mathbf{the}\mathbf{function}$$

Answered by mrW1 last updated on 26/Apr/17

$$\mathbf{y}=\frac{{\mathbf{x}}^3}{3}+2{\mathbf{x}}^2-5\mathbf{x}+7=0$$$$\Rightarrow x=-8.16$$$$$$$$y^{\prime }=x^2+4x-5=0$$$$\Rightarrow (x+5)(x-1)=0$$$$\Rightarrow x=-5$$$$\Rightarrow x=1$$$$$$$$y^{″}=2x+4=0$$$$\Rightarrow x=-2$$

Answered by ajfour last updated on 26/Apr/17

$$\frac{dy}{dx}=x^2+4x-5={(x+2)}^2-9$$$$\Rightarrow \frac{dy}{dx}=0atx=-2\pm 3=-5,1$$$$localmaximaatx=-5$$$$y(-5)=\frac{121}{3}$$$$localminimaatx=1$$$$y\left(1\right)=\frac{13}{3}$$$$\frac{d^{2}y}{{dx}^2}=2x+4$$$$\Rightarrow pointofinflexionatx=-2$$$$y(-2)=\frac{67}{3}$$$$y=0atx=\alpha ;-9<\alpha <-8$$$$cannotevaluate.$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com