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

The tangent of f(x)=x+sqrt(x),\at x=9 is y= 7/6 x+3/2

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

tangent of

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

y = \frac{7}{6} x + \frac{3}{2}

Solution steps

Find the tangent point:

(9, 12)

Find the slope of

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

Compute the slope of

f (x) = x + x

x = 9 : m = \frac{7}{6}

Find the line with slope m=

\frac{7}{6}

and passing through

(9, 12) : y = \frac{7}{6} x + \frac{3}{2}

y = \frac{7}{6} x + \frac{3}{2}

x \to - 1 - lim (\frac{x}{1 + x ^{3}})

\frac{d}{d x} ((8 x^{2} - 5 x) (13 x^{2} + 4))

\frac{\partial}{\partial x} (c e^{- x} sin (4 x))

\int 3 t e^{t} d t

What is the tangent of f(x)=x+sqrt(x),\at x=9 ?
The tangent of f(x)=x+sqrt(x),\at x=9 is y= 7/6 x+3/2