What is the tangent of f(x)=sqrt(3x+7),\at x=3 ?

The tangent of f(x)=sqrt(3x+7),\at x=3 is y= 3/8 x+23/8

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tangent of f(x)=sqrt(3x+7),\at x=3

tangent of

f (x) = 3 x + 7, at x = 3

y = \frac{3}{8} x + \frac{23}{8}

Solution steps

Find the tangent point:

(3, 4)

Find the slope of

f (x) = 3 x + 7 : \frac{df ( x )}{d x} = \frac{3}{2 3 x + 7}

Compute the slope of

f (x) = 3 x + 7

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

Find the line with slope m=

\frac{3}{8}

and passing through

(3, 4) : y = \frac{3}{8} x + \frac{23}{8}

y = \frac{3}{8} x + \frac{23}{8}

(x - 8) \cdot \frac{d y}{d x} + y = (x^{2} - 3)

\frac{\partial}{\partial x} (\frac{4 y}{x ^{3}})

n = 0 \sum \infty \infty

\frac{s + 1}{s ^{2} + 2 s + 10}

\int \frac{2 \cdot x ^{2} - x + 2}{x ^{5} + 2 \cdot x ^{3} + x} d x

What is the tangent of f(x)=sqrt(3x+7),\at x=3 ?
The tangent of f(x)=sqrt(3x+7),\at x=3 is y= 3/8 x+23/8