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Question Number 31670 by gunawan last updated on 12/Mar/18

$$\mathrm{how}\mathrm{many}\mathrm{roots}\mathrm{from}\mathrm{equation}$$$${ae}^x=1+x+\frac{x^2}{2}$$$$froma>0?$$

Answered by mrW2 last updated on 12/Mar/18

$$f\left(x\right)={ae}^x(a>0)$$$$g\left(x\right)=1+x+\frac{x^2}{2}$$$$withx\to -\mathrm{\infty }:$$$$f\left(x\right)\to 0,g\left(x\right)\to +\mathrm{\infty }$$$$i.e.f\left(x\right)<g\left(x\right)$$$$$$$$withx\to +\mathrm{\infty }:$$$$f\left(x\right)\to +\mathrm{\infty },g\left(x\right)\to +\mathrm{\infty }$$$$\frac{f\left(x\right)}{g\left(x\right)}\to +\mathrm{\infty }$$$$i.e.f\left(x\right)>g\left(x\right)$$$$$$$$\Rightarrow between-\mathrm{\infty }and+\mathrm{\infty }thereisone$$$$intersectionpointfromf\left(x\right)andg\left(x\right),$$$$i.e.{ae}^x=1+x+\frac{x^2}{2}hasalwaysoneand$$$$onlyonesolution.$$

Commented by gunawan last updated on 12/Mar/18

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}$$

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