What is the tangent of f(x)=3x+6sqrt(x),(1,9) ?

The tangent of f(x)=3x+6sqrt(x),(1,9) is y=6x+3

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tangent of f(x)=3x+6sqrt(x),(1,9)

tangent of

f (x) = 3 x + 6 x, (1, 9)

y = 6 x + 3

Solution steps

Find the slope of

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

Compute the slope of

f (x) = 3 x + 6 x

(1, 9) : m = 6

Find the line with slope m=

6

and passing through

(1, 9) : y = 6 x + 3

y = 6 x + 3

\int_{5}^{\infty} e^{- \frac{x}{3}} d x

\int \frac{2 x}{9 - x ^{2}} d x

- x + tan (x)

\int_{1}^{6} 2 d x

\int \frac{( 16 - 9 x ^{2} ) ^{3}}{x ^{6}} d x

What is the tangent of f(x)=3x+6sqrt(x),(1,9) ?
The tangent of f(x)=3x+6sqrt(x),(1,9) is y=6x+3