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

how many roots from equation  ae^x =1+x+(x^2 /2)  from a>0 ?

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

Answered by mrW2 last updated on 12/Mar/18

f(x)=ae^x   (a>0)  g(x)=1+x+(x^2 /2)  with x→−∞:  f(x)→0, g(x)→+∞  i.e. f(x)<g(x)    with x→+∞:  f(x)→+∞, g(x)→+∞  ((f(x))/(g(x)))→+∞  i.e. f(x)>g(x)    ⇒between −∞ and +∞ there is one  intersection point from f(x) and g(x),  i.e. ae^x =1+x+(x^2 /2) has always one and  only one solution.

$${f}\left({x}\right)={ae}^{{x}} \:\:\left({a}>\mathrm{0}\right) \\ $$ $${g}\left({x}\right)=\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$ $${with}\:{x}\rightarrow−\infty: \\ $$ $${f}\left({x}\right)\rightarrow\mathrm{0},\:{g}\left({x}\right)\rightarrow+\infty \\ $$ $${i}.{e}.\:{f}\left({x}\right)<{g}\left({x}\right) \\ $$ $$ \\ $$ $${with}\:{x}\rightarrow+\infty: \\ $$ $${f}\left({x}\right)\rightarrow+\infty,\:{g}\left({x}\right)\rightarrow+\infty \\ $$ $$\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\rightarrow+\infty \\ $$ $${i}.{e}.\:{f}\left({x}\right)>{g}\left({x}\right) \\ $$ $$ \\ $$ $$\Rightarrow{between}\:−\infty\:{and}\:+\infty\:{there}\:{is}\:{one} \\ $$ $${intersection}\:{point}\:{from}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right), \\ $$ $${i}.{e}.\:{ae}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{has}\:{always}\:{one}\:{and} \\ $$ $${only}\:{one}\:{solution}. \\ $$

Commented bygunawan last updated on 12/Mar/18

Thank you very much

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

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