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Can anyone help me out how to graph and characterize the function f: [0;PI]->R f(x)=cot x?
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notice that:
f(x) = cot x = 1/tan x
set f'(x) = 0
ie:
d/dx f(x) = -csc^2 x
-1/sin^2 x = 0 -> no solution.
But at the boundaries:
lim as x->0 f(x) = lim x->0 1/tan x -> infinity
So we have a vertical asymptote at x=0.
lim as x->pi f(x) = lim as x->pi 1/tan x -> infinity [since tan x is 0 at x = pi]
So we have a vertical asymptote at x=pi.
Also note that tanget of pi/2 goes off to infinity. So, for 1/tangent:
lim as x->(pi/2) 1/tan x = 0
This is an important point since concavity changes on opposite sides of this.
Now we must check for concavity on both sides of this point using:
d^2/dx^2 f(x) = -csc x * -(csc x)(cot x) = (csc^2 x)(cot x)
Shall I go on?
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