Determine minimum values of a function

Question

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 $0$.

Answer 1

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 $x^*$.

Answer 2

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 $f(x_*)$.