What is the tangent of f(x)=x^3-3x+1,(3,19) ?

The tangent of f(x)=x^3-3x+1,(3,19) is y=24x-53

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tangent of f(x)=x^3-3x+1,(3,19)

tangent of

f (x) = x^{3} - 3 x + 1, (3, 19)

y = 24 x - 53

Solution steps

Find the slope of

f (x) = x^{3} - 3 x + 1 : \frac{df ( x )}{d x} = 3 x^{2} - 3

Compute the slope of

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

(3, 19) : m = 24

Find the line with slope m=

24

and passing through

(3, 19) : y = 24 x - 53

y = 24 x - 53

\int cos^{15} (x) sin (x) d x

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

x \to \infty lim (2 (x))

f (x) = - 6 e^{- 6 x}

- y^{3} \frac{d y}{d x} = x^{5} - 7 x^{2} + 3 x

What is the tangent of f(x)=x^3-3x+1,(3,19) ?
The tangent of f(x)=x^3-3x+1,(3,19) is y=24x-53