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

\frac{2}{r ^{3}}

x \to 0 lim (\frac{1}{ln ( x )})

\frac{2}{s ( s ^{2} + 4 )}

f (x) = - 2 cos (2 x)

\int_{0}^{\frac{π}{2}} 9 sin^{2} (3 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