Finding the Point at Which a Function is a Maximum when the Derivative is a Transcendental Function

Asked Jan 20 '15 at 20:59

Active Jan 20 '15 at 22:00

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Take

$$ f\left(x\right)=-x\left( x\sqrt{4-x^2}-4\arccos\left(\frac{x}{2}\right) \right)\arccos\left(\frac{x^2+d^2-1}{2dx}\right) $$

and try to find the point where $f$ is at a maximum, given $1< x<2$.

I've tried solving $f'\left( x \right)=0$ but the equation is (I think) transcendental in $x$. Mathematica can't solve this either.

If I want to find an approximation to this function (in the form of a second order polynomial) that is accurate at the maximum of $f$, how can I go about it?

edited Jan 20 '15 at 22:00

asked Jan 20 '15 at 20:59

apg

2,797

I'm not sure what values of $d$ you're interested in, but running the code $$Manipulate[ Plot[x ((4 - x^2)^{0.5} - 4 ArcCos[x/2]) ArcCos[(x^2 + d^2 - 1)/( 2 x d)], {x, 0, 3}], {d, 0.000001, 10}]$$ in Mathematica suggests that the maximum is either $x$ arbitrarily close to its upper limit or when $f(x)=0$ – David Simmons Jan 20 '15 at 21:37
the range of d is 2<d<3 – apg Jan 20 '15 at 21:50
OK, doesn't the maximum occur when $f(x)=0$ then? – David Simmons Jan 20 '15 at 21:55
The function clearly takes values greater than 0 – apg Jan 20 '15 at 21:57
You mean f'(x) = 0? – apg Jan 20 '15 at 21:57

Finding the Point at Which a Function is a Maximum when the Derivative is a Transcendental Function

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