What is the tangent of x^3-2x+2(4.58) ?

The tangent of x^3-2x+2(4.58) is y=(3a_{0}^2-2)x-2a_{0}^3+9.16

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tangent of x^3-2x+2(4.58)

tangent of

x^{3} - 2 x + 2 (4.58)

y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 9.16

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 2 a_{0} + 9.16)

Find the slope of

y = x^{3} - 2 x + 2 \cdot 4.58 : \frac{d y}{d x} = 3 x^{2} - 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 = 3 a_{0}^{2} - 2

Find the line with slope m=

3 a_{0}^{2} - 2

and passing through

(a_{0}, a_{0}^{3} - 2 a_{0} + 9.16) : y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 9.16

y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 9.16

\frac{\partial}{\partial x} (3 x - 7 y)

y^{^{'}} cos (x) + y sin (x) = 2 cos^{3} (x) sin (x)

h \to 0 lim (\frac{2 ^{- h} - 1}{h})

arcsin (\frac{x}{3}) + C

\int \frac{54}{( 9 x + 5 ) ^{7}} d x

What is the tangent of x^3-2x+2(4.58) ?
The tangent of x^3-2x+2(4.58) is y=(3a_{0}^2-2)x-2a_{0}^3+9.16