f(x)=ln(tan^2x)f(x)=ln(tan2x)
The Domain: uuu_(k in Z)(-pi/2+kpi, kpi)^^uuu_(k in Z)(kpi,pi/2+kpi)⋃k∈Z(−π2+kπ,kπ)∧⋃k∈Z(kπ,π2+kπ)
f(-x)=ln(tan(-x))^2f(−x)=ln(tan(−x))2
function tanx is odd: tan(-x)=-tanxtan(−x)=−tanx
=>ln((-tanx)^2)=>ln[(-1)^2*(tanx)^2]=>ln(tan^2x)=f(x)⇒ln((−tanx)2)⇒ln[(−1)2⋅(tanx)2]⇒ln(tan2x)=f(x)
Function ln(tan^2x)ln(tan2x) is even
Has periodicity: piπ so I will be graphing only the interval (-pi/2,pi/2)(−π2,π2)
f'(x)=1/tan^2x*2tanx*1/cos^2x
f'(x)=cancel(cos^2x)/sin^2x*2tanx*1/cancel(cos^2x)
f'(x)=(2tanx)/sin^2x
tanx=0hArrx=0
x in (-pi/2,0)hArrf'(x)<0=>f goes down
x in (0,pi/2)hArrf'(x)>0=>f goes up
f''(x)=(2(sinx)^2/(cosx)^2-2tanx*2sinxcosx)/(sinx)^4
f''(x)=(2(sinx)^2/(cosx)^2-2(sinx/cancelcosx)*2sinxcancelcosx)/(sinx)^4
f''(x)=(2/(cosx)^2-2*2)/(sinx)^2
f''(x)=(2(1-2cos^2x))/((cosx)^2(sinx)^2)
cos^2x=1/2=>cosx=+-sqrt2/2
cosx=+-sqrt2/2=>x=+-pi/4
x in (-pi/2,-pi/4)hArrf''(x)>0=>f is convex U
x in (-pi/4,pi/4)hArrf''(x)<0=>f is concave nn
x in (pi/4,pi/2)hArrf''(x)>0=>f is convex U
f(pi/4)=ln(1)=0
Lim_(xrarr0^+)ln(0^+)~~ln(10^())~~-100000(ln10)~~-oo
Now we have everything we need to make a graph.
The Range: RR
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