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)

Compute the slope of

f (x) = 1 + sin (x)

x = - π : m = - 1

Find the line with slope m=

- 1

and passing through

(- π, 1) : y = - x + 1 - π

y = - x + 1 - π

\int x^{3} ln (8 x) d x

x \to - 3 lim (3)

\int \frac{5 ( 4 x + 2 )}{2 ( x ^{2} + 4 x + 5 )} d x

f (x) = \frac{x ^{2}}{( 1 + x )}

n = 1 \sum \infty (1 - \frac{1}{n})^{n}

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