What is the tangent of f(x)=x+sqrt(x),\at x=36 ?

The tangent of f(x)=x+sqrt(x),\at x=36 is y= 13/12 x+3

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tangent of f(x)=x+sqrt(x),\at x=36

tangent of

f (x) = x + x, at x = 36

y = \frac{13}{12} x + 3

Solution steps

Find the tangent point:

(36, 42)

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 = 36 : m = \frac{13}{12}

Find the line with slope m=

\frac{13}{12}

and passing through

(36, 42) : y = \frac{13}{12} x + 3

y = \frac{13}{12} x + 3

\int \frac{( ln ( x ) ) ^{10}}{x} d x

\int \frac{sin ( x )}{1} d x

\frac{d}{d x} (9^{s i n (π x)})

\frac{d}{d x} (- sin (x) + cos (x) + x - \frac{1}{x ^{2}})

f (x) = x + 7, at x = 9

What is the tangent of f(x)=x+sqrt(x),\at x=36 ?
The tangent of f(x)=x+sqrt(x),\at x=36 is y= 13/12 x+3