For what least value of $k$ does the equation:$$e^x=kx^2$$ Have 3 solutions?
Let $f(x)=e^x$ and $g(x)=kx^2$.
For a positive $k$, drawing a rough graph of $f(x)$ and $g(x)$ does show 2 solutions of the equation. But what parameters are we to use while finding the conditions on $k$?
I used the idea that since there exist 3 solutions, there must be a point where the rate of increase of the $g(x)$ is more than that of $f(x)$ and there must also be a point where the rate of increase of $g(x)$ becomes lesser than rate of increase of $f(x)$. But after differentiating, you get another inequality which deals with an exponential function being greater than the linear function. The limiting value of the slope of the line will be the slope of the tangent to the curve $e^x$ which passes through the origin. Again, this answer is in terms of another constant, say, $h$ where the tangent meets the curve. Please advice on how to proceed.
Obviously we must have $k>0$, but then $e^{x}=k x^2$ is equivalent to: $$ e^{x/2} = |x|\sqrt{k}. $$ $e^{x/2}$ is an increasing convex function while $|x|$ is a decreasing convex function on $\mathbb{R}^-$, hence there is exactly one negative solution for any value of $k>0$. There are two positive solution (we cannot have more than two positive solutions, always by convexity) if $\sqrt{k}$ is greater than the slope of the tangent line through the origin to the graph of $e^{x/2}$. The equation of such a tangent line is $y=\frac{e}{2}x$ (it is the tangent line at $x=2$), hence the solution is given by: