What is the tangent of f(x)=sqrt(9-x),\at x=0 ?

The tangent of f(x)=sqrt(9-x),\at x=0 is y=-1/6 x+3

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tangent of f(x)=sqrt(9-x),\at x=0

tangent of

f (x) = 9 - x, at x = 0

y = - \frac{1}{6} x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

f (x) = 9 - x : \frac{df ( x )}{d x} = - \frac{1}{2 9 - 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}{6}

Find the line with slope m=

- \frac{1}{6}

and passing through

(0, 3) : y = - \frac{1}{6} x + 3

y = - \frac{1}{6} x + 3

x \to - 1 lim (3 x^{2} + 2 x - 1)

\frac{6}{s ^{2} ( 7 s + 1 )}

\int 10 x (4 x^{2} - 1)^{9} d x

ln (- 2 x - 5), - 5

\frac{d}{d x} (\frac{e ^{x}}{x})

What is the tangent of f(x)=sqrt(9-x),\at x=0 ?
The tangent of f(x)=sqrt(9-x),\at x=0 is y=-1/6 x+3