What is the tangent of y=x^3-3x+3,(3,21) ?

The tangent of y=x^3-3x+3,(3,21) is y=24x-51

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tangent of y=x^3-3x+3,(3,21)

tangent of

y = x^{3} - 3 x + 3, (3, 21)

y = 24 x - 51

Solution steps

Find the slope of

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

Compute the slope of

y = x^{3} - 3 x + 3

(3, 21) : m = 24

Find the line with slope m=

24

and passing through

(3, 21) : y = 24 x - 51

y = 24 x - 51

\int_{0}^{t} e^{- k t} d t

x \to \frac{π}{2} lim (\frac{sec ( x )}{sec ( 3 x )})

\int (x^{2} - 1)^{3} (2 x) d x

f (x) = 3 + 4 x^{2} - 2 x^{3}, at x = a

(y + 4) y^{^{'}} = 2 x y + 4 y, y (2) = 1

What is the tangent of y=x^3-3x+3,(3,21) ?
The tangent of y=x^3-3x+3,(3,21) is y=24x-51