What is the tangent of f(x)=3x^4-8x^3+2 ?

The tangent of f(x)=3x^4-8x^3+2 is y=(12a_{0}^3-24a_{0}^2)x-9a_{0}^4+16a_{0}^3+2

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tangent of f(x)=3x^4-8x^3+2

tangent of

f (x) = 3 x^{4} - 8 x^{3} + 2

y = (12 a_{0}^{3} - 24 a_{0}^{2}) x - 9 a_{0}^{4} + 16 a_{0}^{3} + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{4} - 8 a_{0}^{3} + 2)

Find the slope of

f (x) = 3 x^{4} - 8 x^{3} + 2 : \frac{df ( x )}{d x} = 12 x^{3} - 24 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} - 24 a_{0}^{2}

Find the line with slope m=

12 a_{0}^{3} - 24 a_{0}^{2}

and passing through

(a_{0}, 3 a_{0}^{4} - 8 a_{0}^{3} + 2) : y = (12 a_{0}^{3} - 24 a_{0}^{2}) x - 9 a_{0}^{4} + 16 a_{0}^{3} + 2

y = (12 a_{0}^{3} - 24 a_{0}^{2}) x - 9 a_{0}^{4} + 16 a_{0}^{3} + 2

\frac{d}{d x} (5 e^{- 2 x} sin (3 x))

(1 + x)^{- \frac{5}{3}}

\frac{\partial}{\partial x} (ln (x y))

\frac{d}{d x} (x^{3} e^{- x})

x \to 9 lim (x + 7)

What is the tangent of f(x)=3x^4-8x^3+2 ?
The tangent of f(x)=3x^4-8x^3+2 is y=(12a_{0}^3-24a_{0}^2)x-9a_{0}^4+16a_{0}^3+2