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

The tangent of f(x)=x^3-3x+2,(2,4) is y=9x-14

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

tangent of

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

y = 9 x - 14

Solution steps

Find the slope of

f (x) = x^{3} - 3 x + 2 : \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 = 9

Find the line with slope m=

9

and passing through

(2, 4) : y = 9 x - 14

y = 9 x - 14

\int sin (π - x) d x

\int - 5 e^{x} d x

\frac{d}{d x} (5 e^{2} + 2 π)

\frac{\partial}{\partial y} (\frac{4 x}{1 + y})

t^{2} \cdot e^{, at}

What is the tangent of f(x)=x^3-3x+2,(2,4) ?
The tangent of f(x)=x^3-3x+2,(2,4) is y=9x-14