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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(6, 2)

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

x = 6 : 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 ^{2}}{d x ^{2}} (\frac{8}{9 t ^{5}})

y = 4 cos (x), y = 6 - 2 cos (x), 0 \leq x \leq 2 π

\frac{d y}{d x} = \frac{cos ( 8 x )}{e ^{8 y}}, y (0) = 1

f (x) = x, at x = 49

\frac{e ^{x}}{x}

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