What is the tangent of f(x)=sqrt(3x),(3,3) ?

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

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tangent of f(x)=sqrt(3x),(3,3)

tangent of

f (x) = 3 x, (3, 3)

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

Solution steps

Find the slope of

f (x) = 3 x : \frac{df ( x )}{d x} = \frac{3}{2 x}

Compute the slope of

f (x) = 3 x

(3, 3) : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

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

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

\frac{2 s + 2}{s ^{2} + 4 s + 5}

\int \frac{1}{2 + 3 x ^{2}} d x

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

\frac{d ^{2}}{d x ^{2}} (2 sec (x))

f (x) = \frac{4 x}{x ^{2} + 1}

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