What is the tangent of y=x^3-4x+3 ?

The tangent of y=x^3-4x+3 is y=(3a_{0}^2-4)x-2a_{0}^3+3

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tangent of y=x^3-4x+3

tangent of

y = x^{3} - 4 x + 3

y = (3 a_{0}^{2} - 4) x - 2 a_{0}^{3} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 4 a_{0} + 3)

Find the slope of

y = x^{3} - 4 x + 3 : \frac{d y}{d x} = 3 x^{2} - 4

Compute the slope of

y = x^{3} - 4 x + 3

x = a_{0} : m = 3 a_{0}^{2} - 4

Find the line with slope m=

3 a_{0}^{2} - 4

and passing through

(a_{0}, a_{0}^{3} - 4 a_{0} + 3) : y = (3 a_{0}^{2} - 4) x - 2 a_{0}^{3} + 3

y = (3 a_{0}^{2} - 4) x - 2 a_{0}^{3} + 3

\int 4 x + 5 d x

x \to 1 lim (\frac{tan ( x )}{x})

\int 100 e^{- 0.01 x} d x

x \to 0 lim (e^{\frac{1}{x}})

\int \frac{2}{x ^{2} - 1} d x

What is the tangent of y=x^3-4x+3 ?
The tangent of y=x^3-4x+3 is y=(3a_{0}^2-4)x-2a_{0}^3+3