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

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

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

tangent of

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

y = - 2 x + 2

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

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

Compute the slope of

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

x = 1 : m = - 2

Find the line with slope m=

- 2

and passing through

(1, 0) : y = - 2 x + 2

y = - 2 x + 2

\frac{d}{d x} (\frac{1}{4} e^{2 x})

\int (sin (3 x) + 3 cos (3 x))^{2} d x

\frac{\partial}{\partial x} (2 x y^{7} + x^{7} y^{9})

\int 6 tan (6 x) d x

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

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