What is the tangent of f(x)=x^3-3x+2,(4,54) ?

The tangent of f(x)=x^3-3x+2,(4,54) is y=45x-126

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tangent of f(x)=x^3-3x+2,(4,54)

tangent of

f (x) = x^{3} - 3 x + 2, (4, 54)

y = 45 x - 126

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 = 45

Find the line with slope m=

45

and passing through

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

y = 45 x - 126

\int_{0}^{2} (15 4 x + 1) d x

x \to 2 lim (\frac{x - 3}{x + 4})

\int x cos (7 π) x^{2} d x

f (r) = \frac{3}{r ^{3}}

n = 1 \sum \infty \frac{x ^{n}}{n 3 ^{n}}

What is the tangent of f(x)=x^3-3x+2,(4,54) ?
The tangent of f(x)=x^3-3x+2,(4,54) is y=45x-126