What is the tangent of f(x)=1+sin(x),\at x=pi ?

The tangent of f(x)=1+sin(x),\at x=pi is y=-x+1+pi

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tangent of f(x)=1+sin(x),\at x=pi

tangent of

f (x) = 1 + sin (x), at x = π

y = - x + 1 + π

Solution steps

Find the tangent point:

(π, 1)

Find the slope of

f (x) = 1 + sin (x) : \frac{df ( x )}{d x} = cos (x)

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 1

Find the line with slope m=

- 1

and passing through

(π, 1) : y = - x + 1 + π

y = - x + 1 + π

f (x) = x^{3} - 74 x^{2} + 40 x + 27490

f (x) = sin^{3} (3 x)

f (t) = cos (2 t)

\frac{d}{d x} (2 \cdot cos (x))

\int (2 x^{4} - 8 x^{3}) e^{- 3 x} d x

What is the tangent of f(x)=1+sin(x),\at x=pi ?
The tangent of f(x)=1+sin(x),\at x=pi is y=-x+1+pi