What is the tangent of f(x)=x^3-3x+1,\at x=4 ?

The tangent of f(x)=x^3-3x+1,\at x=4 is y=45x-127

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tangent of f(x)=x^3-3x+1,\at x=4

tangent of

f (x) = x^{3} - 3 x + 1, at x = 4

y = 45 x - 127

Solution steps

Find the tangent point:

(4, 53)

Find the slope of

f (x) = x^{3} - 3 x + 1 : \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, 53) : y = 45 x - 127

y = 45 x - 127

\frac{d}{d x} (ln (x) + 1)

\int (x^{2} + 7) ln (x) d x

\int \frac{1 + x}{1 - x} d x

\int \frac{x ^{2} + 2 x - 1}{2 x ^{3} + 3 x ^{2} - 2 x} d x

\frac{d}{d x} (x^{2} - 5 x + 4)

What is the tangent of f(x)=x^3-3x+1,\at x=4 ?
The tangent of f(x)=x^3-3x+1,\at x=4 is y=45x-127