Question

Find the relative extrema of the function. Use a graphing utility to confirm your results. (Order your answers from smallest to largest x, then from smallest to largest y.) f(x) = sin(x) sinh(x) − cos(x) cosh(x), −4 ≤ x ≤ 4

Answer 1

Answe$-1 \quad \dfrac{e^{\pi}+e^{-\pi}}{2}$

Step-by-step explanation:

Recall that

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

and that

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

Also note that,

$(\sinh(x))'=\cosh(x)$

and that

$(\cosh(x))'=\sinh(x)$

Now, to find the relative extrema of the function, we first have to compute the derivative of f(x) and find the values of x where the derivative is equal to zero, using the last equations, the linear property of the derivative and the product rule, we get that

$f^{'}(x)=\cos(x)\sinh(x)+\sin(x)\cosh(x)+\sin(x)\cosh(x)-\cos(x)\sinh(x)\\\\=\sin(x)\cosh(x)+\sin(x)\cosh(x)=2\sin(x)\cosh(x)$

Since $\cosh(x) > 0$ for all x, to find the relative extrema of the function in the interval -4≤x≤4, we only have to find the values x, in the given interval, where $\sin(x)=0$. After a inspection of $\sin(x)$ on the given interval, we can see that the extrema of f(x) can be find at $x=-\pi,0,\pi$.

It holds that

$f(0)=-\cos(0)\cosh(0)=(-1) \cdot 1=-1$

and that

$f(-\pi) =f(\pi)=-\cos(\pi)\sinh(\pi)=-(-1)\dfrac{e^{\pi}+e^{-\pi}}{2}=\dfrac{e^{\pi}+e^{-\pi}}{2}$