Question Number 31670 by gunawan last updated on 12/Mar/18 | ||
$$\mathrm{how}\:\mathrm{many}\:\mathrm{roots}\:\mathrm{from}\:\mathrm{equation} \\ $$ $${ae}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$ $${from}\:{a}>\mathrm{0}\:? \\ $$ | ||
Answered by mrW2 last updated on 12/Mar/18 | ||
$${f}\left({x}\right)={ae}^{{x}} \:\:\left({a}>\mathrm{0}\right) \\ $$ $${g}\left({x}\right)=\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$ $${with}\:{x}\rightarrow−\infty: \\ $$ $${f}\left({x}\right)\rightarrow\mathrm{0},\:{g}\left({x}\right)\rightarrow+\infty \\ $$ $${i}.{e}.\:{f}\left({x}\right)<{g}\left({x}\right) \\ $$ $$ \\ $$ $${with}\:{x}\rightarrow+\infty: \\ $$ $${f}\left({x}\right)\rightarrow+\infty,\:{g}\left({x}\right)\rightarrow+\infty \\ $$ $$\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\rightarrow+\infty \\ $$ $${i}.{e}.\:{f}\left({x}\right)>{g}\left({x}\right) \\ $$ $$ \\ $$ $$\Rightarrow{between}\:−\infty\:{and}\:+\infty\:{there}\:{is}\:{one} \\ $$ $${intersection}\:{point}\:{from}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right), \\ $$ $${i}.{e}.\:{ae}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{has}\:{always}\:{one}\:{and} \\ $$ $${only}\:{one}\:{solution}. \\ $$ | ||
Commented bygunawan last updated on 12/Mar/18 | ||
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$ | ||