What is the tangent of f(x)=sqrt(x+7),\at x=2 ?

The tangent of f(x)=sqrt(x+7),\at x=2 is y= 1/6 x+8/3

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tangent of f(x)=sqrt(x+7),\at x=2

tangent of

f (x) = x + 7, at x = 2

y = \frac{1}{6} x + \frac{8}{3}

Solution steps

Find the tangent point:

(2, 3)

Find the slope of

f (x) = x + 7 : \frac{df ( x )}{d x} = \frac{1}{2 x + 7}

Compute the slope of

f (x) = x + 7

x = 2 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

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

y = \frac{1}{6} x + \frac{8}{3}

\frac{d}{d x} (\frac{7 x - 11}{x})

y^{^{'}} = 9 x e^{y}, y (0) = 0

7 - 2 x^{2}, x^{2} + 4

t \to 0 lim (sin (2 t))

\int \frac{1}{x ^{3} x ^{2} - 9} d x

What is the tangent of f(x)=sqrt(x+7),\at x=2 ?
The tangent of f(x)=sqrt(x+7),\at x=2 is y= 1/6 x+8/3