What is the tangent of f(x)=x^3-3x+1,(4,53) ?

The tangent of f(x)=x^3-3x+1,(4,53) is y=45x-127

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tangent of f(x)=x^3-3x+1,(4,53)

tangent of

f (x) = x^{3} - 3 x + 1, (4, 53)

y = 45 x - 127

Solution steps

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

y = ln (5 x^{2} + 9 y^{2})

h \to 0 lim (- 2 x - h + 6)

\int 4 u d u

y = x^{3} - 4 x^{2}, y = 2 x^{2} - 8 x

f (x) = \frac{1}{35} x^{5} + \frac{1}{4} x^{4} + x + 1

What is the tangent of f(x)=x^3-3x+1,(4,53) ?
The tangent of f(x)=x^3-3x+1,(4,53) is y=45x-127