What is the tangent of f(x)= 1/3 x^3-14x^2+6x+1 ?

The tangent of f(x)= 1/3 x^3-14x^2+6x+1 is y=(a_{0}^2-28a_{0}+6)x-2/3 a_{0}^3+14a_{0}^2+1

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tangent of f(x)= 1/3 x^3-14x^2+6x+1

tangent of

f (x) = \frac{1}{3} x^{3} - 14 x^{2} + 6 x + 1

y = (a_{0}^{2} - 28 a_{0} + 6) x - \frac{2}{3} a_{0}^{3} + 14 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{3} a_{0}^{3} - 14 a_{0}^{2} + 6 a_{0} + 1)

Find the slope of

f (x) = \frac{1}{3} x^{3} - 14 x^{2} + 6 x + 1 : \frac{df ( x )}{d x} = x^{2} - 28 x + 6

Compute the slope of

f (x) = \frac{1}{3} x^{3} - 14 x^{2} + 6 x + 1

x = a_{0} : m = a_{0}^{2} - 28 a_{0} + 6

Find the line with slope m=

a_{0}^{2} - 28 a_{0} + 6

and passing through

(a_{0}, \frac{1}{3} a_{0}^{3} - 14 a_{0}^{2} + 6 a_{0} + 1) : y = (a_{0}^{2} - 28 a_{0} + 6) x - \frac{2}{3} a_{0}^{3} + 14 a_{0}^{2} + 1

y = (a_{0}^{2} - 28 a_{0} + 6) x - \frac{2}{3} a_{0}^{3} + 14 a_{0}^{2} + 1

\frac{d}{d x} (- x + \frac{8}{x} + 1)

y^{^{''}} + 10 y^{^{'}} + 24 y = 0

n = 1 \sum \infty 0.5

\int \frac{1}{25 e ^{- 3 x} + e ^{3 x}} d x

(- 3, 5), (2, - 4)

What is the tangent of f(x)= 1/3 x^3-14x^2+6x+1 ?
The tangent of f(x)= 1/3 x^3-14x^2+6x+1 is y=(a_{0}^2-28a_{0}+6)x-2/3 a_{0}^3+14a_{0}^2+1