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 123752 by liberty last updated on 27/Nov/20

$$Letf\left(x\right)bedifferentiablefunction$$$$andg\left(x\right)betheinverseoff\left(x\right).$$$$when\underset{x\to 2}{\lim }\left(\frac{f\left(x\right)-x^3}{x^2-4}\right)=\frac{1}{2}then\frac{dg\left(x\right)}{dx}{\mid }_{x=8}=?$$

Commented by benjo_mathlover last updated on 28/Nov/20

Answered by ajfour last updated on 27/Nov/20

$$sayf\left(x\right)-x^3=2(x-2)$$$$\Rightarrow f\left(x\right)=x^3+2x-4$$$$x=g^3+2g-4$$$$forx=8g^3+2g-12=0$$$$\Rightarrow g=2$$$$1=3g^2\left(\frac{dg}{dX}\right)+2\left(\frac{dg}{dx}\right)$$$$\Rightarrow \left(\frac{dg}{dx}\right){\mid }_{x=8}=\frac{1}{3g^2+2}=\frac{1}{14}.$$$$$$

Answered by bobhans last updated on 27/Nov/20

$$g\left(x\right)=f^{-1}\left(x\right)theng{}^{\prime }\left(x\right)={\left(f^{-1}\right)}^{\prime }\left(x\right)$$$$fromthelimitwegetf\left(2\right)=8$$$$or{f}^{-1}\left(8\right)=2$$$$weknowthat{g}^{\prime }\left(8\right)=\frac{1}{f{}^{\prime }\left(f^{-1}\left(8\right)\right)}...(\bullet )$$$$$$$$consider\underset{x\to 2}{\lim }\frac{f\left(x\right)-x^3}{x^2-4}=\frac{1}{2}$$$$\underset{x\to 2}{\lim }\frac{f{}^{\prime }\left(x\right)-3x^2}{2x}=\frac{1}{2}$$$$\iff f{}^{\prime }\left(2\right)-12=2;f{}^{\prime }\left(2\right)=14$$$$substituteinto(\bullet )weget$$$$g^{\prime }\left(8\right)=\frac{1}{f{}^{\prime }\left(f^{-1}\left(8\right)\right)}=\frac{1}{f{}^{\prime }\left(2\right)}=\frac{1}{14}.$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com