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Find the tangent line equations for the given functions at the given point(s): f(x) = tan x + 9 sin x at (π, 0)

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Answer:


{ \bf{f(x) = \tan x + 9 \sin x }}

For gradient, differentiate f(x):


{ \tt{ (dy)/(dx) = { \sec }^(2)x + 9 \cos x }}

Substitute for x as π:


{ \tt{gradient = { \sec }^(2) \pi + 9 \cos(\pi ) }} \\ { \tt{gradient = - 8 }}

Gradient of tangent = -8


{ \bf{y =mx + b }} \\ { \tt{0 = ( - 8\pi) + b}} \\ { \tt{b = 8\pi}} \\ y - intercept = 8\pi

Equation of tangent:


{ \boxed{ \bf{y = - 8x + 8\pi}}}

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