What is the tangent of f(x)=sqrt(3x+55),\at x=3 ?

The tangent of f(x)=sqrt(3x+55),\at x=3 is y= 3/16 x+119/16

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tangent of f(x)=sqrt(3x+55),\at x=3

tangent of

f (x) = 3 x + 55, at x = 3

y = \frac{3}{16} x + \frac{119}{16}

Solution steps

Find the tangent point:

(3, 8)

Find the slope of

f (x) = 3 x + 55 : \frac{df ( x )}{d x} = \frac{3}{2 3 x + 55}

Compute the slope of

f (x) = 3 x + 55

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

Find the line with slope m=

\frac{3}{16}

and passing through

(3, 8) : y = \frac{3}{16} x + \frac{119}{16}

y = \frac{3}{16} x + \frac{119}{16}

x \to + 2 lim (\frac{g ( x )}{h ( x )})

\int tan^{2} (x) sec^{2} (x) sin (x) d x

y = \frac{( x ^{9} )}{f ( x )}

\frac{\partial}{\partial x} (x e^{2 x} cos (3 x))

\int e^{x} sin (e^{x}) d x

What is the tangent of f(x)=sqrt(3x+55),\at x=3 ?
The tangent of f(x)=sqrt(3x+55),\at x=3 is y= 3/16 x+119/16