Finding the values of constant $a$ so that equation $e^x(3-2\sqrt{x})=a$ has precisely one real solution? [closed]

Find all values of the constant $a$ such that the equation $e^x(3-2\sqrt{x})=a$ has precisely one real solution. $f(x)=e^x(3-2\sqrt{x})$ is defined for $x\ge0$.

I've given it some thought, and I have no idea how to go about solving this problem. Any tips would be helpful.

$$f(x)=e^x(3-2\sqrt{x})$$ Consider that $$f'(x)=e^x\left(3-\left(2\sqrt{x}+\frac1{\sqrt x}\right)\right)$$ and $$f''(x)=e^x\left(3-2\sqrt{x}-\frac2{\sqrt x}+\frac1{2x\sqrt x}\right)$$

The stationary points of $f$ are the points $x$ for which $$f'(x)=0$$ $$3-\left(2\sqrt{x}+\frac1{\sqrt x}\right)=0$$ $$2x-3\sqrt x+1=0$$ $$(2\sqrt x-1)(\sqrt x-1)=0$$ $$\sqrt x=\frac12\text{ or }\sqrt x=1$$ $$x=\frac14\text{ or }x=1$$

Plugging the two stationary points into the second derivative formula yields $$f''\left(\frac14\right)=e^{1/4}\left(3-1-\frac2{1/2}+\frac1{2/8}\right)=2e^{1/4}>0$$ $$f''(1)=e\left(3-2-\frac2{1}+\frac1{2}\right)=-e/2<0$$

Therefore, $\frac14$ is a local minimum point and $1$ is a local maximum point. Now consider that $$f(0)=3$$ $$f\left(\frac14\right)=2e^{1/4}\approx2.5681$$ $$f(1)=e\approx2.7183$$

The graph should roughly look like this (for a smoother version, user Tyma Gaidash has provided a link in a comment)

So, the constants $a$ where $f(x)=a$ has exactly one solution is all $a$ for which $e<a\le3$ or $a<2e^{1/4}$, or restated as a solution set: $$a\in\{x|x<2e^{1/4}\text{ or }e<x\le 3\}$$ As a union of intervals: $$a\in(-\infty,2e^{1/4})\cup(e,3]$$