Thread: Equation has exactly 2 solutions

1. Equation has exactly 2 solutions

$m(x+1)=e^{|x|}$

Find $m\in\mathbb{R}$ so that the equation has exactly two solutions. I really don't know where to start... but for that equation to have any solutions $m(x+1)$ has to be positive and non-zero.

One answer from below is corect:

$A) m\in(1,\infty)$
$B) m\in(-\infty,-e^2)\cup(1,\infty)$
$C) m\in(-\infty,-e^2]\cup[1,\infty)$
$D) m\in(-\infty,-e^2)\cup(0,1)$
$E) m\in{\O}$
F) none of the above

2. Originally Posted by Utherr
$m(x+1)=e^{|x|}$
Well $m(x+1)$ is linear and this looks like $e^x$ for $x\in (0,\infty)$ and then reflected again in the y-axis giving a fat looking parabola.

Originally Posted by Utherr
$m(x+1)=e^{|x|}$
but for that equation to have any solutions $m(x+1)$ has to be positive and non-zero.
do you mean $m$ has to be positive? If so think again, when $m=-9$ you have two solutions

3. Originally Posted by Utherr
$m(x+1)=e^{|x|}$

Find $m\in\mathbb{R}$ so that the equation has exactly two solutions. I really don't know where to start... but for that equation to have any solutions $m(x+1)$ has to be positive and non-zero.

One answer from below is corect:

$A) m\in(1,\infty)$
$B) m\in(-\infty,-e^2)\cup(1,\infty)$
$C) m\in(-\infty,-e^2]\cup[1,\infty)$
$D) m\in(-\infty,-e^2)\cup(0,1)$
$E) m\in{\O}$
F) none of the above
Draw the graph of $y = e^{|x|}$. Now draw possible graphs of $y = m(x + 1)$, taking into account each given option. Come to a conclusion.