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

The tangent of f(x)=sqrt(x),\at x=9 is y= 1/6 x+3/2

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

tangent of

f (x) = x, at x = 9

y = \frac{1}{6} x + \frac{3}{2}

Solution steps

Find the tangent point:

(9, 3)

Find the slope of

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

Compute the slope of

f (x) = x

x = 9 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

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

y = \frac{1}{6} x + \frac{3}{2}

x^{2} - 4 x + 5

y = 2 x + 5

x (x - 4)^{3}

f (x) = \frac{1}{x ^{2}}, at x = 2

(- 7, - 7), (- 6, - 1)

What is the tangent of f(x)=sqrt(x),\at x=9 ?
The tangent of f(x)=sqrt(x),\at x=9 is y= 1/6 x+3/2