What is the tangent of f(x)=sqrt(2-4x),\at x=-1 ?

The tangent of f(x)=sqrt(2-4x),\at x=-1 is y=-sqrt(2/3)x+sqrt(6)-sqrt(2/3)

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tangent of f(x)=sqrt(2-4x),\at x=-1

tangent of

f (x) = 2 - 4 x, at x = - 1

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

Solution steps

Find the tangent point:

(- 1, 6)

Find the slope of

f (x) = 2 - 4 x : \frac{df ( x )}{d x} = - \frac{2}{2 - 4 x}

Compute the slope of

f (x) = 2 - 4 x

x = - 1 : m = - \frac{2}{3}

Find the line with slope m=

- \frac{2}{3}

and passing through

(- 1, 6) : y = - \frac{2}{3} x + 6 - \frac{2}{3}

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

f (x) = n (n^{2} + 3 n - 4)^{7}

\int 56 x e^{4 x^{2}} d x

f (z) = e^{5 \cdot z^{2}}

\frac{10 - 4 s}{( s - 2 ) ^{2}}

\frac{\partial}{\partial x} (- 2 x e^{- 2 x y})

What is the tangent of f(x)=sqrt(2-4x),\at x=-1 ?
The tangent of f(x)=sqrt(2-4x),\at x=-1 is y=-sqrt(2/3)x+sqrt(6)-sqrt(2/3)