What is the tangent of f(x)=sin(x)+1,\at x=-pi/2 ?

The tangent of f(x)=sin(x)+1,\at x=-pi/2 is y=0

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tangent of f(x)=sin(x)+1,\at x=-pi/2

tangent of

f (x) = sin (x) + 1, at x = - \frac{π}{2}

y = 0

Solution steps

Find the tangent point:

(- \frac{π}{2}, 0)

Find the slope of

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

Compute the slope of

f (x) = sin (x) + 1

x = - \frac{π}{2} : m = 0

Find the line with slope m=

0

and passing through

(- \frac{π}{2}, 0) : y = 0

y = 0

f (x) = x^{2} - 4, x = - 4, x = 3

\int \frac{1}{x ^{2} a ^{2} - x ^{2}} d x

x \to - 3 lim (∣ x ∣ - 2)

e^{t}

f (x) = \frac{5 x}{x + 3}, 2, at x = 2

What is the tangent of f(x)=sin(x)+1,\at x=-pi/2 ?
The tangent of f(x)=sin(x)+1,\at x=-pi/2 is y=0