What is the tangent of f(x)=x(1-2x)^3,(1,-1) ?

The tangent of f(x)=x(1-2x)^3,(1,-1) is y=-7x+6

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tangent of f(x)=x(1-2x)^3,(1,-1)

tangent of

f (x) = x (1 - 2 x)^{3}, (1, - 1)

y = - 7 x + 6

Solution steps

Find the slope of

f (x) = x (1 - 2 x)^{3} : \frac{df ( x )}{d x} = (1 - 2 x)^{3} - 6 x (1 - 2 x)^{2}

Compute the slope of

f (x) = x (1 - 2 x)^{3}

(1, - 1) : m = - 7

Find the line with slope m=

- 7

and passing through

(1, - 1) : y = - 7 x + 6

y = - 7 x + 6

\frac{\partial}{\partial w} (\frac{1}{2 v - w})

\int cosh^{3} (x) d x

\int_{- 7}^{7} 49 - x^{2} d x

\frac{\partial}{\partial x} (x^{3} sec^{2} (x y))

\frac{\partial}{\partial x} (13 x + 6 x y^{3})

What is the tangent of f(x)=x(1-2x)^3,(1,-1) ?
The tangent of f(x)=x(1-2x)^3,(1,-1) is y=-7x+6