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This is the equation I'm left with when I took the derivative of a function. I would like to optimise so I'm trying to find a min/max by setting equal to zero. I have been having trouble solving.

$$\frac{2\text{arcosh}(y)}{\sqrt{y^2-1}}+2y-4=0$$

enter image description here

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    Sorry for poor formatting on the bottom of the fraction that is (y^2-1) square rooted – 2017-02-16
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    @JanEerland sorry that is a arcoshy – 2017-02-16
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    can you post us the original Problem? – 2017-02-16
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    Well, you can find the numerical solution: $$y\approx-0.538272389318198812348\space\space\space\bigvee\space\space\space y\approx0.999999999999$$ – 2017-02-16
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    @JanEerland is there a way to actually calculate that like get to that solution through algebra and what not – 2017-02-16
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    I dont think so... – 2017-02-16
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    aha and what about the Constant in this case? – 2017-02-16
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    @Dr.SonnhardGraubner I added picture for context and you will see the original function – 2017-02-16

2 Answers 2

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You want to minimize $$ f(x)=d^2=x^2+(\cosh x-2)^2. $$ Differentiating, we find that $$ f'(x)=2x+2\cosh x\sinh x-4\sinh x, $$ and that $$ f'(0)=0. $$ Now, $$ f''(x)=8\cosh x\sinh^2(x/2)\geq 0 $$ and $f''(x)=0$ if and only if $x=0$. Thus, you have a strictly concave function. Can you conclude from here?

Update If you are not too familiar with convexity, you can argue like this:

$f'(0)=0$, and $f''(x)>0$ if $x>0$. This means that $f'$ is increasing for positive $x$. In particular $f'(x)>f'(0)=0$ if $x>0$. But if $f'(x)>0$ for all $x>0$ it means, in turn, that $f$ is increasing for positive $x$. In particular, $f(x)>f(0)=1$ if $x>0$.

On the other hand $f$ is even ($f(-x)=f(x)$), so $f(x)>f(0)$ also if $x<0$. We conclude that $f$ has a global minimum at $x=0$.

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    I see what you did and I understand it but I don't understand its implication pls forgive I'm only senior in high school can you explain a bit please – 2017-02-16
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    That is fine. I updated my answer. If it is not clearer now, don't hesitate to ask again. – 2017-02-16
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    That helped a lot thanks man/woman go to be pc ya know – 2017-02-16
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    Nicely written (+1) – 2017-02-16
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$y=\cosh(x)$ is a convex function and the curvature of its graph at $(x,\cosh(x))$ equals $\cosh(x)$.
This is enough to conclude that the shortest path for Bob for reaching the catenary is to go towards its vertex: see catenary evolute. As an alternative, you may notice that the squared distance of Bob from $(x,\cosh(x))$ is given by $$ x^2+(2-\cosh(x))^2 = 1+\sum_{m\geq 2}\frac{4^m-8}{(2m)!}x^{2m} $$ that is an increasing function of $x^2$, since $\frac{4^m-8}{(2m)!}>0$ for any $m\geq 2$.