What is the tangent of 6x^2-x^3 ?

The tangent of 6x^2-x^3 is y=(12a_{0}-3a_{0}^2)x+2a_{0}^3-6a_{0}^2

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tangent of 6x^2-x^3

tangent of

6 x^{2} - x^{3}

y = (12 a_{0} - 3 a_{0}^{2}) x + 2 a_{0}^{3} - 6 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 6 a_{0}^{2} - a_{0}^{3})

Find the slope of

y = 6 x^{2} - x^{3} : \frac{d y}{d x} = 12 x - 3 x^{2}

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 = 12 a_{0} - 3 a_{0}^{2}

Find the line with slope m=

12 a_{0} - 3 a_{0}^{2}

and passing through

(a_{0}, 6 a_{0}^{2} - a_{0}^{3}) : y = (12 a_{0} - 3 a_{0}^{2}) x + 2 a_{0}^{3} - 6 a_{0}^{2}

y = (12 a_{0} - 3 a_{0}^{2}) x + 2 a_{0}^{3} - 6 a_{0}^{2}

\int \frac{5.45}{( 1 + 1.9 x ) ^{0.9}} d x

f (x) = ln (4 - x^{6})

\int x sin (\frac{nπ x}{20}) d x

x \to 2 - lim (\frac{x ^{2} - 4}{x - 2})

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

What is the tangent of 6x^2-x^3 ?
The tangent of 6x^2-x^3 is y=(12a_{0}-3a_{0}^2)x+2a_{0}^3-6a_{0}^2