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$$\rm{ Calculate: }\quad \displaystyle \sup_{x\in (0,2)}\left|\cosh\left(x^2-x-1\right)\right|$$

My proof:

$$f(x)=\cosh\left(x^2-x-1\right)$$ $$f'(x)=\left(2x-1 \right)\sinh\left(x^2-x-1\right) $$ $$f'(x)=0 \iff x=\dfrac{1}{2} \rm{ or } \sinh\left(x^2-x-1\right)=0 $$ note that :

$$\sinh(x)=\dfrac{e^{x}-e^{-x}}{2}$$

  • $\sinh\left(x^2-x-1\right)=0$

\begin{align*} \sinh\left(x^2-x-1\right)=0 &\iff \dfrac{e^{x^2-x-1}-e^{-x^2+x+1}}{2}=0\\ &\iff e^{x^2-x-1}=e^{-x^2+x+1}\\ &\iff x=\dfrac{1\pm \sqrt{5}}{2} \end{align*}

\begin{array}{c|ccccccc} x & 0 & & \frac{1}{2} & & \frac{1+\sqrt{5}}{2} & & 2 \\ \hline f'(x) & & + & 0 & - & 0 & + & \\ \hline f(x) & \cosh(1) & \nearrow &\cosh\left(\dfrac{5}{4}\right) & \searrow & \cosh(0) & \nearrow &\cosh(1) \end{array}

then

$$\displaystyle \sup_{x\in (0,2)}\left|\cosh\left(x^2-x-1\right)\right|=\max\{ |f(0)|;|f(1/2)|,|f( \frac{1+\sqrt{5}}{2})|,|f(2)|\}=\cosh\left(\dfrac{5}{4}\right)$$

  • Is my proof correct ?
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    $\begingroup$ i got also $$\cosh\left(\frac{5}{4}\right)$$ and the equal sign holds if $$x=\frac{1}{2},x=1$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 18 '16 at 17:11
  • $\begingroup$ @Dr.SonnhardGraubner $x=1?$ $\endgroup$ – zhw. Aug 18 '16 at 18:23
  • $\begingroup$ $$y=1$$ it was a typo,sorry $\endgroup$ – Dr. Sonnhard Graubner Aug 18 '16 at 18:26
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Alternate proof: Let $p(x) = x^2 - x - 1.$ Verify that $p([0,2]) = [-5/4,1].$ Now $\cosh x$ is an even function. Therefore

$$\sup_{x\in [0,2]} \cosh (p(x))= \sup_{y\in [-5/4,1]} \cosh y = \sup_{y\in [0,5/4]} \cosh y.$$

On the last interval $\cosh y$ is increasing, therefore the desired answer is $\cosh (5/4).$ (Note that $\cosh$ is always positive, so your absolute values are not needed.)

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