What is the tangent of f(x)=arccos(x)+x,\at x=0 ?

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

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tangent of f(x)=arccos(x)+x,\at x=0

tangent of

f (x) = arccos (x) + x, at x = 0

y = \frac{π}{2}

Solution steps

Find the tangent point:

(0, \frac{π}{2})

Find the slope of

f (x) = arccos (x) + x : \frac{df ( x )}{d x} = - \frac{1}{1 - x ^{2}} + 1

Compute the slope of

f (x) = arccos (x) + x

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, \frac{π}{2}) : y = \frac{π}{2}

y = \frac{π}{2}

t \frac{d y}{d t} + 18 y = \frac{1}{t}

f (x) = \frac{2 x - 1}{x + 3}, at x = 3

\int - sin (t) d t

\frac{\partial ^{2}}{\partial x \partial y} (ln (3 x^{2} - y^{3} - 3))

\frac{d}{d x} (9 e^{- x} + e^{4 x})

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