Finding if there is a maximum or minimum on a curve?

Question

My apologies for being very brief with this question, the reason for this is because I don't know where to start.

The question is as follows: A curve has the equation $\lambda \cosh(x) + \sinh(x)$, where $\lambda$ is a constant.

Determine the coordinates of the turning point of the curve in the case when . Is this a maximum or minimum?

If anyone can shed some light on how to arrive at a solution to this question I would be very appreciative. Thank you

Answer 1

Let $f(x) = \lambda \cosh(x)+\sinh(x)$.

Its derivative is :

$f'(x) = \lambda \sinh(x)+\cosh(x)$

We are looking for the zeroes of $f'$:

\begin{align}{f'(x) = 0 \iff \\ \lambda \sinh(x)+\cosh(x)=0 \iff \\ \lambda(e^x-e^{-x})=-e^x-e^{-x } \iff \\ \lambda(e^{2x}-1)=-e^{2x}-1} \iff\\ e^{2x}(\lambda+1)=-1+\lambda \\ \end{align} This has a solution if $\lambda ≠ -1$ and $a=\frac{\lambda - 1}{\lambda+1}>0$. In such a case, you get $b=\frac{1}{2}\ln(a)$ as unique solution.

Then you compute $f''$ and check that $f''(b)≠0$ where $b$ is zero of $f'$. More precisely, if $f''(b)>0$ then $x=b$ is a local minimum of $f$, and if $f''(b)<0$ then $x=b$ is a local maximum of $f$.

Assume that $a>0$ as previously (so that the unique zero $b$ of $f'$ does exist). We have $f''(x)=f(x)$ and $f''(b)>0$ if $\lambda>1$, $f''(b)<0$ if $\lambda<-1$.

The coordinates are then :

If $\lambda>1$, then $f$ a unique minimum at $x_0=\frac{1}{2}\ln\left(\frac{\lambda - 1}{\lambda+1}\right)$, assuming that $a=\frac{\lambda - 1}{\lambda+1}>0$ (otherwise $f$ has no extremum). If $\lambda<-1$, then $f$ a unique maximum at $x_0=\frac{1}{2}\ln\left(\frac{\lambda - 1}{\lambda+1}\right)$, assuming that $a=\frac{\lambda - 1}{\lambda+1}>0$ (otherwise $f$ has no extremum). You can prove that $f(x_0) = \text{sign}(\lambda)\sqrt{\lambda^2 - 1}$.