What is the tangent of f(x)=x-x^3,(1,0) ?

The tangent of f(x)=x-x^3,(1,0) is y=-2x+2

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tangent of f(x)=x-x^3,(1,0)

tangent of

f (x) = x - x^{3}, (1, 0)

y = - 2 x + 2

Solution steps

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}

(1, 0) : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 2

x^{2} + 2 x y - y^{2} + x = 27, (4, 7)

\frac{d y}{d x} = \frac{y ^{3}}{1 - 2 x y ^{2}}, y (0) = 1

\frac{\partial}{\partial x} (sin (z^{2}) cos (x y))

- cos (θ)

g (x) = (6 x^{4} + 9)^{2} (x^{3} - 4 x)^{3}

What is the tangent of f(x)=x-x^3,(1,0) ?
The tangent of f(x)=x-x^3,(1,0) is y=-2x+2