What is the tangent of y= 1/3 x^3-3x^2+5x ?

The tangent of y= 1/3 x^3-3x^2+5x is y=(a_{0}^2-6a_{0}+5)x-2/3 a_{0}^3+3a_{0}^2

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tangent of y= 1/3 x^3-3x^2+5x

tangent of

y = \frac{1}{3} x^{3} - 3 x^{2} + 5 x

y = (a_{0}^{2} - 6 a_{0} + 5) x - \frac{2}{3} a_{0}^{3} + 3 a_{0}^{2}

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} - 3 a_{0}^{2} + 5 a_{0})

Find the slope of

y = \frac{1}{3} x^{3} - 3 x^{2} + 5 x : \frac{d y}{d x} = x^{2} - 6 x + 5

Compute the slope of

y = \frac{1}{3} x^{3} - 3 x^{2} + 5 x

x = a_{0} : m = a_{0}^{2} - 6 a_{0} + 5

Find the line with slope m=

a_{0}^{2} - 6 a_{0} + 5

and passing through

(a_{0}, \frac{1}{3} a_{0}^{3} - 3 a_{0}^{2} + 5 a_{0}) : y = (a_{0}^{2} - 6 a_{0} + 5) x - \frac{2}{3} a_{0}^{3} + 3 a_{0}^{2}

y = (a_{0}^{2} - 6 a_{0} + 5) x - \frac{2}{3} a_{0}^{3} + 3 a_{0}^{2}

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

x \to 0 lim (\frac{( 4 + x ) ^{- 1} - 4 ^{- 1}}{x})

g (t) = \frac{5}{t}

2^{2}

What is the tangent of y= 1/3 x^3-3x^2+5x ?
The tangent of y= 1/3 x^3-3x^2+5x is y=(a_{0}^2-6a_{0}+5)x-2/3 a_{0}^3+3a_{0}^2