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

The tangent of f(x)=-3x^3-2x^2-1 is y=(-9a_{0}^2-4a_{0})x+6a_{0}^3+2a_{0}^2-1

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tangent of f(x)=-3x^3-2x^2-1

tangent of

f (x) = - 3 x^{3} - 2 x^{2} - 1

y = (- 9 a_{0}^{2} - 4 a_{0}) x + 6 a_{0}^{3} + 2 a_{0}^{2} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 3 a_{0}^{3} - 2 a_{0}^{2} - 1)

Find the slope of

f (x) = - 3 x^{3} - 2 x^{2} - 1 : \frac{df ( x )}{d x} = - 9 x^{2} - 4 x

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 = - 9 a_{0}^{2} - 4 a_{0}

Find the line with slope m=

- 9 a_{0}^{2} - 4 a_{0}

and passing through

(a_{0}, - 3 a_{0}^{3} - 2 a_{0}^{2} - 1) : y = (- 9 a_{0}^{2} - 4 a_{0}) x + 6 a_{0}^{3} + 2 a_{0}^{2} - 1

y = (- 9 a_{0}^{2} - 4 a_{0}) x + 6 a_{0}^{3} + 2 a_{0}^{2} - 1

\int_{1}^{32} x^{- \frac{9}{5}} d x

(- 9 x^{2} + 4)^{2}

\int \frac{2 x ^{4} + 3 x}{2 x ^{2}} d x

\int \frac{1}{1 + t} d t

\frac{4 x}{x ^{2} + 4}

What is the tangent of f(x)=-3x^3-2x^2-1 ?
The tangent of f(x)=-3x^3-2x^2-1 is y=(-9a_{0}^2-4a_{0})x+6a_{0}^3+2a_{0}^2-1