The range is on the interval [-1-sqrt(3),1]
The points are on the graphs x=sqrt(1+2cos(y)) and x=-sqrt(1+2cos(y)) which form a vertical arrangement of ovals. (blue and green on the graph)
The new function can then be split into
f(x)=sqrt(1+2cos(x))-cos(x)
g(x)=-sqrt(1+2cos(x))-cos(x)
(Which form the horizontal red and black heart shapes on the graph.)
https://www.desmos.com/calculator/vdhdfeewga
I'll spare the formulas, but using the derivative: f'(x)=0 when sin(x)=0 or cos(x)=0 from which the maximum is f(pi/2)=1
g'(x)=0 when sin(x)=0 from which the minimum is f(0)=-1-sqrt(3)
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Posted by Jer
on 2021-09-20 12:41:52 |