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

5 x y

f (x) = (5 x + 2)^{\frac{1}{3}}, at x = 2

\int_{0}^{1} (x^{2} + 4) e^{- x} d x

t \cdot e^{t}

\int_{0}^{\frac{π}{2}} cos^{2} (x) 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