What is the tangent of f(x)=sqrt(x+2) ?

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

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tangent of f(x)=sqrt(x+2)

tangent of

f (x) = x + 2

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} + 2)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{2 a _{0} + 2}

Find the line with slope m=

\frac{1}{2 a _{0} + 2}

and passing through

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

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

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