What is the tangent of f(x)=sqrt(3-6x) ?

The tangent of f(x)=sqrt(3-6x) is y=-3/(sqrt(3-6a_{0))}x+sqrt(3-6a_{0)}+(3a_{0})/(sqrt(3-6a_{0))}

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tangent of f(x)=sqrt(3-6x)

tangent of

f (x) = 3 - 6 x

y = - \frac{3}{3 - 6 a _{0}} x + 3 - 6 a_{0} + \frac{3 a _{0}}{3 - 6 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 - 6 a_{0})

Find the slope of

f (x) = 3 - 6 x : \frac{df ( x )}{d x} = - \frac{3}{3 - 6 x}

Compute the slope of

f (x) = 3 - 6 x

x = a_{0} : m = - \frac{3}{3 - 6 a _{0}}

Find the line with slope m=

- \frac{3}{3 - 6 a _{0}}

and passing through

(a_{0}, 3 - 6 a_{0}) : y = - \frac{3}{3 - 6 a _{0}} x + 3 - 6 a_{0} + \frac{3 a _{0}}{3 - 6 a _{0}}

y = - \frac{3}{3 - 6 a _{0}} x + 3 - 6 a_{0} + \frac{3 a _{0}}{3 - 6 a _{0}}

x \to 1 lim (- x + 2)

\frac{\partial}{\partial x} (x + 3 \cdot ln (x) - y^{2})

\frac{d}{d x} (a x^{2} + x)

\int_{5}^{6} \frac{1}{x ^{2} - 1} d x

\int_{- 2}^{3} \frac{18}{x ^{4}} d x

What is the tangent of f(x)=sqrt(3-6x) ?
The tangent of f(x)=sqrt(3-6x) is y=-3/(sqrt(3-6a_{0))}x+sqrt(3-6a_{0)}+(3a_{0})/(sqrt(3-6a_{0))}