What is the tangent of f(x)=sqrt(x),(16,4) ?

The tangent of f(x)=sqrt(x),(16,4) is y= 1/8 x+2

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tangent of f(x)=sqrt(x),(16,4)

tangent of

f (x) = x, (16, 4)

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

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x

(16, 4) : 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

x \to 5 lim (2 x^{2} - 5 x + 3)

y = x + 8

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

f (x) = \frac{x ^{3} - 3 x ^{2} + 4}{x ^{2}}

5 - x^{2}, sin (x), - 2.38468, 2.02521

What is the tangent of f(x)=sqrt(x),(16,4) ?
The tangent of f(x)=sqrt(x),(16,4) is y= 1/8 x+2