What is the tangent of f(x)=sqrt(x-3) ?

The tangent of f(x)=sqrt(x-3) is y= 1/(2sqrt(a_{0)-3)}x+sqrt(a_{0)-3}-(a_{0})/(2sqrt(a_{0)-3)}

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tangent of f(x)=sqrt(x-3)

tangent of

f (x) = x - 3

y = \frac{1}{2 a _{0} - 3} x + a_{0} - 3 - \frac{a _{0}}{2 a _{0} - 3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} - 3)

Find the slope of

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

Compute the slope of

f (x) = x - 3

x = a_{0} : m = \frac{1}{2 a _{0} - 3}

Find the line with slope m=

\frac{1}{2 a _{0} - 3}

and passing through

(a_{0}, a_{0} - 3) : y = \frac{1}{2 a _{0} - 3} x + a_{0} - 3 - \frac{a _{0}}{2 a _{0} - 3}

y = \frac{1}{2 a _{0} - 3} x + a_{0} - 3 - \frac{a _{0}}{2 a _{0} - 3}

\int_{- 4}^{2} f (x) + 2 x + 5 d x

\frac{1}{2} cos (2 t)

\int \frac{1}{u a ^{2} - u ^{2}} d u

\frac{\partial}{\partial z} (cos (x yz))

\frac{d y}{d t} = 1.4 y

What is the tangent of f(x)=sqrt(x-3) ?
The tangent of f(x)=sqrt(x-3) is y= 1/(2sqrt(a_{0)-3)}x+sqrt(a_{0)-3}-(a_{0})/(2sqrt(a_{0)-3)}