Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 38

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.