What is the tangent of f(x)=sqrt(x+8),\at x=8 ?

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

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tangent of f(x)=sqrt(x+8),\at x=8

tangent of

f (x) = x + 8, at x = 8

y = \frac{1}{8} x + 3

Solution steps

Find the tangent point:

(8, 4)

Find the slope of

f (x) = x + 8 : \frac{df ( x )}{d x} = \frac{1}{2 x + 8}

Compute the slope of

f (x) = x + 8

x = 8 : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(8, 4) : y = \frac{1}{8} x + 3

y = \frac{1}{8} x + 3

y^{^{''}} + 4 y^{^{'}} = 2 e^{2 x}

\int - k

4 x + \frac{5}{x}

\frac{\partial}{\partial x} (6 - x^{2} + x y - 4 y^{2})

x^{2} + 1, x + 1

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