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

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

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

tangent of

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

y = 9 x - 14

Solution steps

Find the tangent point:

(2, 4)

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

\frac{\partial}{\partial y} (2 e^{2 z} x y - 2 y)

\frac{1}{x ^{3}}

\frac{d}{d t} (cos (t) - sin (t))

\int + \frac{e ^{2 x}}{1 + e ^{2 x}} d x

f (x) = 4 x^{2} - 4 x + 1, at x = 1

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