What is the tangent of f(x)=sqrt(3x+1),\at x=5 ?

The tangent of f(x)=sqrt(3x+1),\at x=5 is y= 3/8 x+17/8

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tangent of f(x)=sqrt(3x+1),\at x=5

tangent of

f (x) = 3 x + 1, at x = 5

y = \frac{3}{8} x + \frac{17}{8}

Solution steps

Find the tangent point:

(5, 4)

Find the slope of

f (x) = 3 x + 1 : \frac{df ( x )}{d x} = \frac{3}{2 3 x + 1}

Compute the slope of

f (x) = 3 x + 1

x = 5 : m = \frac{3}{8}

Find the line with slope m=

\frac{3}{8}

and passing through

(5, 4) : y = \frac{3}{8} x + \frac{17}{8}

y = \frac{3}{8} x + \frac{17}{8}

\frac{d}{d x} (2 e^{6 x})

\int x 1 + x^{4} d x

\frac{d}{d x} (sin^{2} (x - 3 y))

\int e^{x} 3 - e^{x} d x

What is the tangent of f(x)=sqrt(3x+1),\at x=5 ?
The tangent of f(x)=sqrt(3x+1),\at x=5 is y= 3/8 x+17/8