What is the tangent of f(x)=sqrt(x-2),\at x=3 ?

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

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tangent of f(x)=sqrt(x-2),\at x=3

tangent of

f (x) = x - 2, at x = 3

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

Solution steps

Find the tangent point:

(3, 1)

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 = 3 : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

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

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

cos (2 x) + sin (x)

y^{^{''}} - 6 y^{^{'}} + 9 y = 0

\int_{0}^{L} x d x

y^{^{'}} + 4 y = x^{2}

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