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

The tangent of f(x)=x^3-3x^2+2x-2 is y=(3a_{0}^2-6a_{0}+2)x-2a_{0}^3+3a_{0}^2-2

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

tangent of

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

y = (3 a_{0}^{2} - 6 a_{0} + 2) x - 2 a_{0}^{3} + 3 a_{0}^{2} - 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 3 a_{0}^{2} + 2 a_{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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 3 a_{0}^{2} - 6 a_{0} + 2

Find the line with slope m=

3 a_{0}^{2} - 6 a_{0} + 2

and passing through

(a_{0}, a_{0}^{3} - 3 a_{0}^{2} + 2 a_{0} - 2) : y = (3 a_{0}^{2} - 6 a_{0} + 2) x - 2 a_{0}^{3} + 3 a_{0}^{2} - 2

y = (3 a_{0}^{2} - 6 a_{0} + 2) x - 2 a_{0}^{3} + 3 a_{0}^{2} - 2

\int_{2}^{9} \frac{4}{8 x - 9} d x

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

f (x) = 7 + 4 x^{2} - 2 x^{3}, (1, 9)

\int \frac{2 x}{- 5 - 6 x ^{2} - x ^{4}} d x

\frac{7 x}{8 - tan ( x )}

What is the tangent of f(x)=x^3-3x^2+2x-2 ?
The tangent of f(x)=x^3-3x^2+2x-2 is y=(3a_{0}^2-6a_{0}+2)x-2a_{0}^3+3a_{0}^2-2