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

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

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

tangent of

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

y = 2 x - 6

Solution steps

Find the tangent point:

(2, - 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 = 2 : m = 2

Find the line with slope m=

2

and passing through

(2, - 2) : y = 2 x - 6

y = 2 x - 6

\int 2 x^{2} + 2 x d x

\int e^{- 0.5 t} d t

\frac{\partial}{\partial x} (22 cos (x))

\frac{\partial}{\partial y} (\frac{y}{x})

\int - 25 cos (x) d x

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