What is the tangent of f(x)=sqrt(6-x),\at x=3 ?

The tangent of f(x)=sqrt(6-x),\at x=3 is y=-1/(2sqrt(3))x+sqrt(3)+(sqrt(3))/2

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tangent of f(x)=sqrt(6-x),\at x=3

tangent of

f (x) = 6 - x, at x = 3

y = - \frac{1}{2 3} x + 3 + \frac{3}{2}

Solution steps

Find the tangent point:

(3, 3)

Find the slope of

f (x) = 6 - x : \frac{df ( x )}{d x} = - \frac{1}{2 6 - 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 3}

Find the line with slope m=

- \frac{1}{2 3}

and passing through

(3, 3) : y = - \frac{1}{2 3} x + 3 + \frac{3}{2}

y = - \frac{1}{2 3} x + 3 + \frac{3}{2}

\int x^{2} e^{- 6 x} d x

e x^{2}

\int_{- 2}^{2} (x^{2} - 4) d x

\int \frac{( 2 x + 1 )}{x ^{2} + 4 x + 5} d x

x \to 0 lim (\frac{( 1 + x ) ^{4} - 1}{x})

What is the tangent of f(x)=sqrt(6-x),\at x=3 ?
The tangent of f(x)=sqrt(6-x),\at x=3 is y=-1/(2sqrt(3))x+sqrt(3)+(sqrt(3))/2