What is the tangent of f(x)=sqrt(x),\at x=16 ?

The tangent of f(x)=sqrt(x),\at x=16 is y= 1/8 x+2

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tangent of f(x)=sqrt(x),\at x=16

tangent of

f (x) = x, at x = 16

y = \frac{1}{8} x + 2

Solution steps

Find the tangent point:

(16, 4)

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}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(16, 4) : y = \frac{1}{8} x + 2

y = \frac{1}{8} x + 2

\frac{d}{d x} (\frac{1}{1 + x ^{12}})

\frac{d}{d x} (1 - e^{- 8 x})

2 x + 5 x^{- 1}

\int_{1}^{4} (4 x^{2} - 2 x) d x

(\frac{3 y ^{2} - x ^{2}}{y ^{5}}) \frac{d y}{d x} + \frac{x}{2 y ^{4}} = 0

What is the tangent of f(x)=sqrt(x),\at x=16 ?
The tangent of f(x)=sqrt(x),\at x=16 is y= 1/8 x+2