What is the tangent of f(x)=x^3-3x^2+2x-2,\at x=0 ?

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

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tangent of f(x)=x^3-3x^2+2x-2,\at x=0

tangent of

f (x) = x^{3} - 3 x^{2} + 2 x - 2, at x = 0

y = 2 x - 2

Solution steps

Find the tangent point:

(0, - 2)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x - 2

\frac{\partial}{\partial y} (\frac{3 x - 4 y}{3 x + 4 y})

x \to 9 lim (cos (\frac{1}{x - 9}))

f (θ) = 2 tan (θ)

\int 4 sin (\frac{x}{2}) + 8 d x

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