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 94214 by abony1303 last updated on 17/May/20

$$Findf\left(0\right)whenapolynomialf\left(x\right)satisfies$$$$\underset{x\to 1}{\lim }\frac{f\left(x\right)}{x^2-1}=2$$$$\underset{x\to -1}{\lim }\frac{f\left(x\right)}{x^2-1}=2$$$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{f\left(x\right)}{x^4}=1plsHelp!$$

Commented by john santu last updated on 17/May/20

$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{\mathrm{f}\left(\mathrm{x}\right)}{{\mathrm{x}}^4}=??$$

Commented by john santu last updated on 17/May/20

$$\mathrm{f}\left(x\right)=(x^2-1)({ax}^2+b)$$$$$$

Commented by abony1303 last updated on 17/May/20

$$changed$$

Commented by i jagooll last updated on 17/May/20

$$f\left(x\right)=(x^2-1)({ax}^2+bx+c)$$$$\left(1\right)\underset{x\to 1}{\lim }\frac{(x^2-1)({ax}^2+bx+c)}{x^2-1}=2$$$$\Rightarrow a+b+c=2$$$$\left(2\right)\underset{x\to -1}{\lim }\frac{(x^2-1)({ax}^2+bx+c)}{x^2-1}=2$$$$\Rightarrow a-b+c=2$$$$\left(3\right)\underset{x\to +\mathrm{\infty }}{\lim }\frac{f\left(x\right)}{x^4}=1\Rightarrow a=1$$$$eq\left(1\right)+\left(2\right)\Rightarrow 2a+2c=4\Rightarrow c=1and$$$$b=0.thenf\left(x\right)=(x^2-1)(x^2+1)$$$$f\left(0\right)=-1$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com