# How do you graph y=ln(tan^2 x)?

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#### Spiegazione:

f(x)=ln(tan^2x)
The Dominio: uuu_(k in Z)(-pi/2+kpi, kpi)^^uuu_(k in Z)(kpi,pi/2+kpi)

f(-x)=ln(tan(-x))^2
function tanx is odd: tan(-x)=-tanx
=>ln((-tanx)^2)=>ln[(-1)^2*(tanx)^2]=>ln(tan^2x)=f(x)

Funzione ln(tan^2x) è anche
Has periodicity: pi so I will be graphing only the interval (-pi/2,pi/2)

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