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

The tangent of f(x)=sqrt(x+1),\at x=8 is y= 1/6 x+5/3

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tangent of f(x)=sqrt(x+1),\at x=8

tangent of

f (x) = x + 1, at x = 8

y = \frac{1}{6} x + \frac{5}{3}

Solution steps

Find the tangent point:

(8, 3)

Find the slope of

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

Compute the slope of

f (x) = x + 1

x = 8 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(8, 3) : y = \frac{1}{6} x + \frac{5}{3}

y = \frac{1}{6} x + \frac{5}{3}

\int (x^{2} + 2 x)^{3} d x

t sin (t)

\frac{d}{d x} (\frac{2}{( 3 x ^{2} + 7 ) ^{2}})

\frac{\partial}{\partial x} (e^{x y} sin (z))

\frac{d}{d x} (4 x^{2 x})

What is the tangent of f(x)=sqrt(x+1),\at x=8 ?
The tangent of f(x)=sqrt(x+1),\at x=8 is y= 1/6 x+5/3