Minimising Function, derivative with exponentials

Question

First of all, I apologise for not giving a more descriptive title. I really do not know how to word it. I'll go straight into the meat of the question.

If a function $$h(x)=\frac{e^x-1}{x^5}$$ is to be minimised, then you go about finding the first derivative and solving that for zero to find the critical points.

I've done that and get the following $$h'(x)=\frac{xe^x-5e^x+5}{x^6}$$ and to find the critical points we then get $xe^x-5e^x+5=0$.

I am either being extremely stupid or there is no way which you can use to isolate $x$ and so can't solve via a "straightforward" method.

It should follow into a transcendental equation in the form $x=g(x)$.

My research suggests there is a solution in terms of Lambert W functions, however this has not yet be taught in my university course, and checking with the lecturer we do not need them, i.e. "straightforward" method...

What exactly am I missing here?

Thanks in advance.

Answer 1

You can use the fact that the function $g(x)=xe^x-5e^x+5$ is monotone on the interval [4,6] and that, by the Bolzano's theorem, g(4) < 0 and g(6)>0 then exists an unique $\alpha \in ]4,6[$ where $g(\alpha)=0$. for this value $\alpha$ you can calculated the value of $f(\alpha)$ and make the variations of the function $f$. The exact value of $\alpha$, which is near 5, is not fundamental.

$g(\alpha)=\alpha e^\alpha-5e^\alpha+5 =0$ so $\alpha e^\alpha = 5(e^\alpha -1)$ and $f(\alpha)= \frac{\alpha e^\alpha}{5 \alpha^5} =\frac{e^\alpha}{5\alpha^4}$.

Answer 2

As I commented, $h'(x)$ cancels close to $x=5$. Since $$h'(x)=\frac{e^x (x-5)+5}{x^6}$$ excluding $x=0$, we have to look at $$g(x)=e^x (x-5)+5$$ and we can notice that $g(4)=5-e^4\approx -49.5982$ and $g(5)=5$. A quick estimation of the solution of $g(x)=0$ can be made using Taylor expansion around $x=5$ and get $$g(x)=5+e^5 (x-5)+e^5 (x-5)^2+O\left((x-5)^3\right)$$ Using the first order gives $x_*=\frac{5 \left(e^5-1\right)}{e^5}\approx 4.96631$ So $$h(x_*)=\frac{e^{x_*}-1}{x_*^5}\approx\frac{e^{5}-1}{5^5}=\frac{e^5-1}{3125}\approx 0.0471722$$ while the exact value would be $\approx 0.0471666$.

The second derivative test confirms that this is a minimum.