What is the tangent of f(x)=3sqrt(x),\at x=16 ?

The tangent of f(x)=3sqrt(x),\at x=16 is y= 3/8 x+6

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tangent of f(x)=3sqrt(x),\at x=16

tangent of

f (x) = 3 x, at x = 16

y = \frac{3}{8} x + 6

Solution steps

Find the tangent point:

(16, 12)

Find the slope of

f (x) = 3 x : \frac{df ( x )}{d x} = \frac{3}{2 x}

Compute the slope of

f (x) = 3 x

x = 16 : m = \frac{3}{8}

Find the line with slope m=

\frac{3}{8}

and passing through

(16, 12) : y = \frac{3}{8} x + 6

y = \frac{3}{8} x + 6

\int sec^{3} (2 x) d x

\int_{0}^{4} 4 y - y^{2} d y

\frac{400}{( s ( s ^{2} + 200 ) )}

x^{2} e^{3 x}

\frac{d}{d x} ((arctan (2 x))^{2})

What is the tangent of f(x)=3sqrt(x),\at x=16 ?
The tangent of f(x)=3sqrt(x),\at x=16 is y= 3/8 x+6