What is the tangent of y=x^3-3x+2(2.4) ?

The tangent of y=x^3-3x+2(2.4) is y=(3a_{0}^2-3)x-2a_{0}^3+4.8

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tangent of y=x^3-3x+2(2.4)

tangent of

y = x^{3} - 3 x + 2 (2.4)

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 4.8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 3 a_{0} + 4.8)

Find the slope of

y = x^{3} - 3 x + 2 \cdot 2.4 : \frac{d y}{d x} = 3 x^{2} - 3

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

Find the line with slope m=

3 a_{0}^{2} - 3

and passing through

(a_{0}, a_{0}^{3} - 3 a_{0} + 4.8) : y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 4.8

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 4.8

\int tan^{5} (π x) sec^{4} (π x) d x

x^{4} + 7 x^{2} - x

\int 2 x^{3} (3 x^{4} - 5) d x

\int_{0}^{2 π} sin^{2} (2 x) d x

\int \frac{1 + 3 t ^{5}}{2 t} d t

What is the tangent of y=x^3-3x+2(2.4) ?
The tangent of y=x^3-3x+2(2.4) is y=(3a_{0}^2-3)x-2a_{0}^3+4.8