What is the tangent of f(x)=12sqrt(x),\at x=9 ?

The tangent of f(x)=12sqrt(x),\at x=9 is y=2x+18

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tangent of f(x)=12sqrt(x),\at x=9

tangent of

f (x) = 12 x, at x = 9

y = 2 x + 18

Solution steps

Find the tangent point:

(9, 36)

Find the slope of

f (x) = 12 x : \frac{df ( x )}{d x} = \frac{6}{x}

Compute the slope of

f (x) = 12 x

x = 9 : m = 2

Find the line with slope m=

2

and passing through

(9, 36) : y = 2 x + 18

y = 2 x + 18

(t^{2} + 4) y^{^{'}} + 2 t y = t^{2} (t^{2} + 4)

y = x^{2} + 3, (4, 19)

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

\frac{d}{d y} (\frac{2 x ^{2} - x y + 2 y ^{2}}{x - y})

x^{2} + 2 y^{2} - 2 x - 8 y + 6 = 0 (0.3)

What is the tangent of f(x)=12sqrt(x),\at x=9 ?
The tangent of f(x)=12sqrt(x),\at x=9 is y=2x+18