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

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

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tangent of f(x)=x^3-3x+2\at (2.4)

tangent of

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

f (x) = x^{3} - 3 x + 22.4 : \frac{df ( x )}{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

x = y, x = \frac{y}{2}

x \to \infty lim (2 x^{2} - 3)

x^{2} - 4, x = - 4, x = 3, y = 0

\int_{0}^{1} (3 x^{2} - 4 x + 1) d x

y^{^{''}} + \frac{1}{4} y = 0

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