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

The tangent of f(x)=x^3-3x+2,\at x=4 is y=45x-126

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

tangent of

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

y = 45 x - 126

Solution steps

Find the tangent point:

(4, 54)

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 = 45

Find the line with slope m=

45

and passing through

(4, 54) : y = 45 x - 126

y = 45 x - 126

y = \frac{1}{x}, y = 2 x - 1, 0, 3

x \to \infty lim (4 cos (3 x))

\int 0.55 t^{2} + 1.575 t^{4} d t

\int \frac{x}{x ^{- \frac{5}{2}}} d x

\int 3 x^{2} - 3 y^{2} d y

What is the tangent of f(x)=x^3-3x+2,\at x=4 ?
The tangent of f(x)=x^3-3x+2,\at x=4 is y=45x-126