What is the tangent of f(x)=sqrt(x-2),(6,2) ?

The tangent of f(x)=sqrt(x-2),(6,2) is y= 1/4 x+1/2

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tangent of f(x)=sqrt(x-2),(6,2)

tangent of

f (x) = x - 2, (6, 2)

y = \frac{1}{4} x + \frac{1}{2}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x - 2

(6, 2) : m = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(6, 2) : y = \frac{1}{4} x + \frac{1}{2}

y = \frac{1}{4} x + \frac{1}{2}

\frac{d y}{d x} 7 + e^{- 5 x} y = 0

f (x) = arctan (x^{2})

(ln (x^{2} - 10 x - 10)), at x = 11

(x^{2} + 1) (\frac{d y}{d x}) + 3 x (y - 1) = 0, y (0) = 8

x \to - 3 lim (x x + 3)

What is the tangent of f(x)=sqrt(x-2),(6,2) ?
The tangent of f(x)=sqrt(x-2),(6,2) is y= 1/4 x+1/2