Find all values of $a$ for which equation $ e^{x^2} = ax$ always has exactly two roots.

Question

How shall I solve the following question? Please help!

"Find all values of '$a$' for which equation $ e^{x^2} = ax$ always has exactly two roots."

Answer 1

Edited since it was answered and I had a slightly different way in mind.

We want the positive $b$ value such that $bx$ intersects $e^{x^2}$ only once. Then for $|a| >b$ we have $ax$ intersecting $e^{x^2}$ twice. The positive value $b$ will yield an intersection point at a positive $x$ value $x_0$

The tangent line at each point has a slope of $f'(x) = 2xf(x)$. So we want to solve $f'(x_0)x_0 = f(x_0)$ which implies that $2x_0^2 = 1$ and $x_0 = \frac{1}{\sqrt{2}}$

Solving $bx_0 = e^{x_0^2}$ we get that $b\frac{1}{\sqrt{2}} =\sqrt{e} $

Answer 2

I would start by getting a sense of when the equation has exactly one root.

Letting $f(x)=e^{x^2},$ we have that $$f'(x)=e^{x^2}\cdot\frac{d}{dx}\left[x^2\right]=2x\cdot f(x).$$ Since $f(x)=a_0x$ having exactly one root means that $y=a_0x$ is a tangent line to some point of $y=f(x),$ then we should figure out what the tangent lines of $y=f(x)$ can look like.

Pick an arbitrary $x_0.$ We know that the tangent line to $y=f(x)$ at $x_0$ has slope $f'(x_0)=2x_0\cdot f(x_0),$ and passes through the point $\bigl(x_0,f(x_0)\bigr).$ Thus, by point-slope form, we have $$y-f(x_0)=2x_0\cdot f(x_0)\cdot(x-x_0)\\y-f(x_0)=2x_0\cdot f(x_0)\cdot x-2x_0^2\cdot f(x_0)\\y=2x_0\cdot f(x_0)\cdot x+(1-2x_0^2)\cdot f(x_0).$$ What can we conclude about $x_0$ if this has the form $y=a_0x$? There are actually two closely-related values it can have. What then can $a_0$ be?

Now, note that if $|a|<|a_0|,$ then $f(x)=ax$ has no solutions. It remains to show that $f(x)=ax$ has two solutions whenever $|a|>|a_0|,$ which you can do by showing that $f$ is convex (since it has a positive second derivative).

Answer 3

This is one of those cases where the picture tells the story. First draw the graphs of these functions, and then check that

$(1).\ f(x)=e^{x^2}$ is a convex even function, decreases from $-\infty$ to its absolute minimum at $x=0,$ and from there it increases to $\infty.$ These facts are easily checked by taking and interpreting derivatives.

$(2).\ g(x)=ax$ is a line passing through the origin, with slope $m=a.$ As $a$ increases, the graph of $g$ rises, until it just touches the graph of $f$. Here is the calculation: We want $x_0$ such that $e^{x_0^2}=ax_0$ and $2x_0e^{x_0^2}=a.$ That is, $2x_0^2=1$ or $x_0=\frac{\pm 1}{\sqrt 2},\ $ from which $a=\pm \sqrt2e^{1/2}.$

$(3)$. Now suppose $a>\sqrt2e^{1/2},\ x> 0$. Then, $h(x)=ax-e^{x^2}=0\Leftrightarrow \ln a+\ln x=x^2$. Set $\phi(x)=x^2-\ln x -\ln a.$ We want to show that $\phi$ has exactly two roots on $(0,\infty)$ and $a> \sqrt2e^{1/2}$. Now, since $\phi'(x)=2x-\frac{1}{x},\ \phi$ has one and only one critical point at $x=\frac{1}{\sqrt 2},$ which is a local minimum. And since $\phi(1/\sqrt 2)<0$, we see that the graph of $\phi$ decreases from $\infty$ to $\phi(1/\sqrt 2)$, turns around, and from there increases to $\infty.$ By the intermediate value theorem, the graph of $\phi$ intersects the $x$ axis exactly twice, as desired.

$(4).$ Now take $x<0$ and $a<-\sqrt2e^{1/2}$, and repeat the analysis in $(3)$ to show that the claim is true for $a<-\sqrt2e^{1/2}$ or $a>\sqrt2e^{1/2}.$