What are the values of $m$ $m$ for which inequality is true? $(m + 2) e^{- 2 x} + 2 (m + 2) e^{- x} + m > 0$ $(m+2)e^(-2x)+2(m+2)e^-x+m>0$ .

Question

870 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-26T22:38:43

$m > 2 (\frac{1}{{(e^{- x} + 1)}^{2}} - 1)$ $m > 2(1/(e^-x+1)^2-1)$

Calling $y = e^{- x}$ $y = e^-x$ and assuming $m \neq - 2$ $m ne -2$ we have

$y^{2} + 2 y + \frac{m}{m + 2} > 0$ $y^2+2y+m/(m+2) > 0$ or

${(y + 1)}^{2} - 1 + \frac{m}{m + 2} > 0$ $(y+1)^2-1 +m/(m+2) > 0$

or

$\frac{m}{m + 2} > 1 - {(y + 1)}^{2}$ $m/(m+2) > 1-(y+1)^2$

or

$m > 2 \frac{1 - {(y + 1)}^{2}}{{(y + 1)}^{2}}$ $m > 2(1-(y+1)^2)/(y+1)^2$

or

$m > 2 (\frac{1}{{(y + 1)}^{2}} - 1) = 2 (\frac{1}{{(e^{- x} + 1)}^{2}} - 1)$ $m > 2(1/(y+1)^2-1)=2(1/(e^-x+1)^2-1)$

Attached a plot showing in light blue the feasible region in the plane $x \times m$ $x xx m$

enter image source here

What are the values of mm for which inequality is true? (m+2)e−2x+2(m+2)e−x+m>0(m+2)e^(-2x)+2(m+2)e^-x+m>0.