`f(x) = xsinh(x-1) - cosh(x-1)` Find any relative extrema of the function.

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This function is defined on entire `RR` and is infinitely differentiable. The necessary condition of extremum for such a function is `f'(x) = 0.`

The derivative of `f(x)` is

`sinh(x-1) + xcosh(x-1) - sinh(x-1) =xcosh(x-1).` 

The only solution of the equation `f'(x) = 0` is `x = 0,` because `cosh(x) gt 0` for all `x.` Moreover, this fact gives us that `f'(x)` is positive for positive `x` and negative for negative `x,` so `f(x)` decreases on `(-oo, 0)` and increases on `(0, +oo).`

Therefore `x = 0` is the point where a local minimum is reached. The value of the function at this point is `-cosh(-1) = -(e + 1/e)/2 approx -1.543.`

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