What is the tangent of f(x)=(x^3)/3+2x^2+13x-5 ?

The tangent of f(x)=(x^3)/3+2x^2+13x-5 is y=(a_{0}^2+4a_{0}+13)x-a_{0}^3-2a_{0}^2+(a_{0}^3)/3-5

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tangent of f(x)=(x^3)/3+2x^2+13x-5

tangent of

f (x) = \frac{x ^{3}}{3} + 2 x^{2} + 13 x - 5

y = (a_{0}^{2} + 4 a_{0} + 13) x - a_{0}^{3} - 2 a_{0}^{2} + \frac{a _{0}^{3}}{3} - 5

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} + 13 a_{0} - 5)

Find the slope of

f (x) = \frac{x ^{3}}{3} + 2 x^{2} + 13 x - 5 : \frac{df ( x )}{d x} = x^{2} + 4 x + 13

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} + 13

Find the line with slope m=

a_{0}^{2} + 4 a_{0} + 13

and passing through

(a_{0}, \frac{a _{0}^{3}}{3} + 2 a_{0}^{2} + 13 a_{0} - 5) : y = (a_{0}^{2} + 4 a_{0} + 13) x - a_{0}^{3} - 2 a_{0}^{2} + \frac{a _{0}^{3}}{3} - 5

y = (a_{0}^{2} + 4 a_{0} + 13) x - a_{0}^{3} - 2 a_{0}^{2} + \frac{a _{0}^{3}}{3} - 5

n = 1 \sum \infty \frac{n}{ln ( 2 n )}

x^{3} - 2 x + 2, at x = 3, y = 21

\frac{\partial}{\partial x} (\frac{8 x + 5 y}{z})

\frac{\partial}{\partial x} (2 (\frac{y}{x})^{0.5})

- \frac{y}{( x ^{2} + y ^{2} )} d x + \frac{x}{x ^{2} + y ^{2}} d y = 0

What is the tangent of f(x)=(x^3)/3+2x^2+13x-5 ?
The tangent of f(x)=(x^3)/3+2x^2+13x-5 is y=(a_{0}^2+4a_{0}+13)x-a_{0}^3-2a_{0}^2+(a_{0}^3)/3-5