What is the tangent of f(x)=sqrt(x+9),\at x=0 ?

The tangent of f(x)=sqrt(x+9),\at x=0 is y= 1/6 x+3

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tangent of f(x)=sqrt(x+9),\at x=0

tangent of

f (x) = x + 9, at x = 0

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

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

f (x) = x + 9 : \frac{df ( x )}{d x} = \frac{1}{2 x + 9}

Compute the slope of

f (x) = x + 9

x = 0 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

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

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

\int (7 x^{2} + 3 x - 4) d x

x \to 2 lim (3 x^{2} + 4 x + 10)

\int_{π}^{2 π} cos (x) d x

4 cot (\frac{x}{7})

\frac{d}{d x} ((8 (2 x)^{4} + 5)^{3})

What is the tangent of f(x)=sqrt(x+9),\at x=0 ?
The tangent of f(x)=sqrt(x+9),\at x=0 is y= 1/6 x+3