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nevsk [136]
3 years ago
14

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

Mathematics
1 answer:
xenn [34]3 years ago
6 0

Answer:

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

For gradient, differentiate f(x):

{ \tt{ \frac{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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