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

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

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

tangent of

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

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

Find the slope of

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

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} - 3

Find the line with slope m=

3 a_{0}^{2} - 3

and passing through

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

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

\int \frac{4}{16 x ^{2} + 1} d x

\int 2 e^{2 t + 5} d t

\frac{d}{d x} ((ln (3) + e^{x^{e}})^{e})

y^{^{''}} + y^{^{'}} - 2 y = 9 e^{t} - 6 t

y^{^{'}} - sin (x) y = 3 sin (x)

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