What is the tangent of f(x)=x^3-9x,\at x=3 ?

The tangent of f(x)=x^3-9x,\at x=3 is y=18x-54

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tangent of f(x)=x^3-9x,\at x=3

tangent of

f (x) = x^{3} - 9 x, at x = 3

y = 18 x - 54

Solution steps

Find the tangent point:

(3, 0)

Find the slope of

f (x) = x^{3} - 9 x : \frac{df ( x )}{d x} = 3 x^{2} - 9

Compute the slope of

f (x) = x^{3} - 9 x

x = 3 : m = 18

Find the line with slope m=

18

and passing through

(3, 0) : y = 18 x - 54

y = 18 x - 54

\int (1 + \frac{1}{x})^{2} \frac{1}{x ^{2}} d x

\frac{d}{d x} (arccos (3 x + 1))

\int 3^{x} x d x

f (0) = \frac{x}{x ^{2} + 4}

\int_{0}^{1} (\frac{x ^{3} + 1}{x + 1}) d x

What is the tangent of f(x)=x^3-9x,\at x=3 ?
The tangent of f(x)=x^3-9x,\at x=3 is y=18x-54