What is the tangent of (x^3)/3+2x^2+4x-1 ?

The tangent of (x^3)/3+2x^2+4x-1 is y=(a_{0}^2+4a_{0}+4)x-a_{0}^3-2a_{0}^2+(a_{0}^3)/3-1

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tangent of (x^3)/3+2x^2+4x-1

tangent of

\frac{x ^{3}}{3} + 2 x^{2} + 4 x - 1

y = (a_{0}^{2} + 4 a_{0} + 4) x - a_{0}^{3} - 2 a_{0}^{2} + \frac{a _{0}^{3}}{3} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0}^{3}}{3} + 2 a_{0}^{2} + 4 a_{0} - 1)

Find the slope of

y = \frac{x ^{3}}{3} + 2 x^{2} + 4 x - 1 : \frac{d y}{d x} = x^{2} + 4 x + 4

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 = a_{0}^{2} + 4 a_{0} + 4

Find the line with slope m=

a_{0}^{2} + 4 a_{0} + 4

and passing through

(a_{0}, \frac{a _{0}^{3}}{3} + 2 a_{0}^{2} + 4 a_{0} - 1) : y = (a_{0}^{2} + 4 a_{0} + 4) x - a_{0}^{3} - 2 a_{0}^{2} + \frac{a _{0}^{3}}{3} - 1

y = (a_{0}^{2} + 4 a_{0} + 4) x - a_{0}^{3} - 2 a_{0}^{2} + \frac{a _{0}^{3}}{3} - 1

x \to 0 lim (6 sin (x) - 1)

\frac{lo g _{10} ( 2 - x )}{x + 3}

\frac{d}{d x} ((4 x - 5)^{2})

n = 5 \sum \infty \frac{1}{n ^{2} - 1}

x \to 0 lim (6 x + 3)

What is the tangent of (x^3)/3+2x^2+4x-1 ?
The tangent of (x^3)/3+2x^2+4x-1 is y=(a_{0}^2+4a_{0}+4)x-a_{0}^3-2a_{0}^2+(a_{0}^3)/3-1