What is the tangent of f(x)=18sqrt(x),\at x=9 ?

The tangent of f(x)=18sqrt(x),\at x=9 is y=3x+27

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tangent of f(x)=18sqrt(x),\at x=9

tangent of

f (x) = 18 x, at x = 9

y = 3 x + 27

Solution steps

Find the tangent point:

(9, 54)

Find the slope of

f (x) = 18 x : \frac{df ( x )}{d x} = \frac{9}{x}

Compute the slope of

f (x) = 18 x

x = 9 : m = 3

Find the line with slope m=

3

and passing through

(9, 54) : y = 3 x + 27

y = 3 x + 27

n = 2 \sum \infty \frac{7}{4 ^{n}}

x \to \infty lim (e^{x} x)

\int \frac{3}{x + 4} d x

x \to 4 lim ((2 x - 3)^{2})

\int_{0}^{2} \frac{1}{x ^{2} - 4} d x

What is the tangent of f(x)=18sqrt(x),\at x=9 ?
The tangent of f(x)=18sqrt(x),\at x=9 is y=3x+27