What is the tangent of f(x)=sqrt(x+2),\at (x=4) ?

The tangent of f(x)=sqrt(x+2),\at (x=4) is y= 1/(2sqrt(6))x+sqrt(6)-sqrt(2/3)

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tangent of f(x)=sqrt(x+2),\at (x=4)

tangent of

f (x) = x + 2, at (x = 4)

y = \frac{1}{2 6} x + 6 - \frac{2}{3}

Solution steps

Find the tangent point:

(4, 6)

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 6}

Find the line with slope m=

\frac{1}{2 6}

and passing through

(4, 6) : y = \frac{1}{2 6} x + 6 - \frac{2}{3}

y = \frac{1}{2 6} x + 6 - \frac{2}{3}

What is the tangent of f(x)=sqrt(x+2),\at (x=4) ?
The tangent of f(x)=sqrt(x+2),\at (x=4) is y= 1/(2sqrt(6))x+sqrt(6)-sqrt(2/3)