Determine minimum values of a function Find the $x$ values that minimize the function $f(x)$.
$$
f(x) = e^x + \frac{1}{m}(x-4)^2
$$
I know I can determine the minimum values when the derivative of the function is $0$.
So  I have calculated the derivative
$$
f'(x) = e^x + \frac{2x}{m} - \frac{8}{m}
$$
But I do not know how to reflect $x$ when $f'(x)$ is $0$.
 A: You can observe that the derivative is monotonically increasing and that is negative when $x\to-\infty$ and positive when $x\to+\infty$. You can then conclude on the existence of an $x^*$ such that $f'(x^*)=0$. Moreover, since $f''(x^*)>0$, this is a minimum.
The equation
$$f'(x^*) = e^{x^*} + \frac{2x^*}{m} - \frac{8}{m}=0$$
is a transcendental equation and so there will no closed-form solution for $x^*$ in terms of elementary functions. We can, however, express the real solution in terms of the Lambert-W function that expresses the solutions of equations of the form $xe^x=c$.
First let $y=2x^*/m+8/m$ and then $x^*=(my+8)/2$. Considering then $e^{-x^*}f'(x^*)=0$ and the previous change of variables yields
$$1+e^{-(my+8)/2}y=0.$$
This expression can be reformulated as
$$-\dfrac{my}{2}e^{-my/2}=\dfrac{m}{2}e^4.$$
Invoking now the Lambert W function, we get that
$$y = \dfrac{-2}{m}W_0(-me^4/2),$$
where $W_0$ is the principal branch of the Lambert W function. Finally,
$$x^*=-W(-me^4/2)+4,$$
which gives an explicit expression for $x^*$.
A: Just for your curiosity.
@KBS gave the only explicit solution : $f(x)$ is minimum when $x=x_*=4-W\left(\frac{e^4 m}{2}\right)$ where $W(.)$ is Lambert function.
This makes
$$f(x_*)=\frac 1 m W\left(\frac{e^4 m}{2}\right) \left(2+W\left(\frac{e^4 m}{2}\right)\right)$$
In year $2007$, Hoorfar and Hassani gave in this paper the following tight bounds $\forall x \geq e$
$$\log (x)-\log (\log (x))+\frac 12\frac{\log (\log (x))}{\log (x)}\leq W(x)\leq \log (x)-\log (\log (x))+\frac e {e-1}\frac{\log (\log (x))}{\log (x)}$$
This gives you a quick way to bound the value of $f(x_*)$.
