What is the tangent of f(x)=sqrt(x),\at x=36 ?

The tangent of f(x)=sqrt(x),\at x=36 is y= 1/12 x+3

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tangent of f(x)=sqrt(x),\at x=36

tangent of

f (x) = x, at x = 36

y = \frac{1}{12} x + 3

Solution steps

Find the tangent point:

(36, 6)

Find the slope of

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

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}{12}

Find the line with slope m=

\frac{1}{12}

and passing through

(36, 6) : y = \frac{1}{12} x + 3

y = \frac{1}{12} x + 3

What is the tangent of f(x)=sqrt(x),\at x=36 ?
The tangent of f(x)=sqrt(x),\at x=36 is y= 1/12 x+3