Suppose we have a function of the form $$f(x) = \frac{J_1(kx)}{x}$$ where $J_1$ is a Bessel function of the first kind, and k is some constant, and we would like to to find $x$ such that this function is maximised, how would one go about doing this?
I am not very familiar with Bessel functions, and the definitions of Bessel functions that I have seen look unsuitable for straightforward differentiation. Is there a way to go about finding other global and local maxima/minima of this type of function? Is it possible to do this without resorting to numerical methods?
Critical points of $f$
$J_1$ is a solution of
$$x^2J_1''(x)+xJ_1'(x)+(x^2-1)J_1(x)=0.$$ So we have
$$k^2x^2J_1''(kx)+kxJ_1'(kx)+(k^2x^2-1)J_1(kx)=0.$$ Now it is
$$f'(x)=\dfrac{kJ_1'(kx)}{x}-\dfrac{J_1(kx)}{x^2}=0\iff kxJ_1'(kx)=J_1(kx).$$ Thus we have
$$k^2x^2J_1''(kx)+k^2x^2J_1(kx)=0$$ and $$kxJ_1''(kx)+J_1'(kx)=0.$$ So, it must be
$$(k^2x^2-1)J_1(kx)=0.$$ That is, $$x=\dfrac{\pm 1}{k}$$ or $J_1(kx)=0.$