What is the tangent of f(x)=sqrt(x-5),\at x=6 ?

The tangent of f(x)=sqrt(x-5),\at x=6 is y= 1/2 x-2

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tangent of f(x)=sqrt(x-5),\at x=6

tangent of

f (x) = x - 5, at x = 6

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

Solution steps

Find the tangent point:

(6, 1)

Find the slope of

f (x) = x - 5 : \frac{df ( x )}{d x} = \frac{1}{2 x - 5}

Compute the slope of

f (x) = x - 5

x = 6 : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

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

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

x = 7 y^{3}, x = 18 y^{2}

\int g e^{- c x} d x

f (x) = ln (x^{2} - 2 x + 1), at x = 2

\frac{( x ^{2} )}{2 + x}

\int e^{- 0.06 x} d x

What is the tangent of f(x)=sqrt(x-5),\at x=6 ?
The tangent of f(x)=sqrt(x-5),\at x=6 is y= 1/2 x-2