Find the equation of the tangent line for f(x)=4tan(x) at x=π/6?
回答 (2)
2015-09-26 2:31 am
✔ 最佳答案
f'x) = 4sec^2(x)f'(pi/6) = 4*4/3 = 16/3
y = f(pi/6) = 4*sqrt(3)/3
Equation in point slope form
y- 4*sqrt(3)/3 = (16/3)(x-pi/6)
2015-09-26 2:48 am
tangent:y - f (xo)= f'(xo) (x - xo)
xo=TT/6
f(xo)=f (TT/6)= 4tan (TT/6)= 4/ sqrt (3)=
(4sqrt 3 )/3.
f'(x)= 4 [tan^2 (x)+1]
f'(TT/6)= 4[ tan^2 ( TT/6)+1]=
= 4 [(1/(sqrt 3)^2+1]=
=4 (1/3+1)=4 (4/3)=16/3.
Conclusione:
Tangent:
y-4sqrt(3)/3=16/3 (x-TT/6)
y=(16/3) x - 8TT/9+4sqrt 3/.
xo=TT/6
f(xo)=f (TT/6)= 4tan (TT/6)= 4/ sqrt (3)=
(4sqrt 3 )/3.
f'(x)= 4 [tan^2 (x)+1]
f'(TT/6)= 4[ tan^2 ( TT/6)+1]=
= 4 [(1/(sqrt 3)^2+1]=
=4 (1/3+1)=4 (4/3)=16/3.
Conclusione:
Tangent:
y-4sqrt(3)/3=16/3 (x-TT/6)
y=(16/3) x - 8TT/9+4sqrt 3/.
收錄日期: 2021-04-21 14:26:07
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