What is the tangent of f(x)=6sqrt(x),\at x=9 ?

The tangent of f(x)=6sqrt(x),\at x=9 is y=x+9

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tangent of f(x)=6sqrt(x),\at x=9

tangent of

f (x) = 6 x, at x = 9

y = x + 9

Solution steps

Find the tangent point:

(9, 18)

Find the slope of

f (x) = 6 x : \frac{df ( x )}{d x} = \frac{3}{x}

Compute the slope of

f (x) = 6 x

x = 9 : m = 1

Find the line with slope m=

1

and passing through

(9, 18) : y = x + 9

y = x + 9

\int 2 e^{x} x d x

\int \frac{( 6 x - 9 )}{( x ^{2} - 3 x + 5 ) ^{3}} d x

\frac{d}{d x} (- \frac{1}{3} cos (3 x))

y = ln (x^{5})

f (x) = (\frac{x + 1}{x - 1})^{7}

What is the tangent of f(x)=6sqrt(x),\at x=9 ?
The tangent of f(x)=6sqrt(x),\at x=9 is y=x+9