What is the tangent of f(x)=sqrt(1-x),\at x=0 ?

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

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tangent of f(x)=sqrt(1-x),\at x=0

tangent of

f (x) = 1 - x, at x = 0

y = - \frac{1}{2} x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Find the line with slope m=

- \frac{1}{2}

and passing through

(0, 1) : y = - \frac{1}{2} x + 1

y = - \frac{1}{2} x + 1

x \to 0 lim (5)

\frac{d y}{d x} = e^{y} (x^{2} - x)

\frac{d}{d x} (\frac{2 x}{9 - tan ( x )})

n = 1 \sum \infty 4 (\frac{2}{3})^{n}

\frac{d}{d x} (- 2 sin (4 x))

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