how to find the inverse

Question

gamma

using this equation

$1-\sqrt{1-x^2/c^2}$

where c = 1 and x= 0.0 - 1.0 the speed of c

for example

$1-\sqrt{1-.886^2/1^2}$ = y = 0.5363147619

gives me the y values on the graph. How do I find the inverse? How do find for x inputting the values of y?

Thank you.

Answer 1

If $f(x)=1-\frac{1}{\frac{1}{\sqrt{1-x^2/c^2}}}$, then $f(x)=1-\sqrt{1-x^2/c^2}.$

Hence $\sqrt{1-x^2/c^2}=1-f(x)$, thus $1-x^2/c^2=1-2f(x)+f(x)^2.$ Can you proceed ?

Answer 2

Here is the graph for $0\le x\le 1$

For the inverse, note that $$(x,y)\in f\iff (y,x)\in f^{-1}$$therefore the inverse would be found from $$x=1-\sqrt{1-y^2}$$from which we obtain $$f^{-1}(x)=\sqrt{2x-x^2}$$