What is the tangent of f(x)=x^3-11x,\at x=3 ?

The tangent of f(x)=x^3-11x,\at x=3 is y=16x-54

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tangent of f(x)=x^3-11x,\at x=3

tangent of

f (x) = x^{3} - 11 x, at x = 3

y = 16 x - 54

Solution steps

Find the tangent point:

(3, - 6)

Find the slope of

f (x) = x^{3} - 11 x : \frac{df ( x )}{d x} = 3 x^{2} - 11

Compute the slope of

f (x) = x^{3} - 11 x

x = 3 : m = 16

Find the line with slope m=

16

and passing through

(3, - 6) : y = 16 x - 54

y = 16 x - 54

n = 2 \sum \infty (\frac{3}{4})^{n}

\frac{d}{d x} ((4 x^{5} - 2 x) (3 x + e^{x}))

\frac{7}{t}

x \to \infty lim (2 x^{3} - 5 x - 4 x^{3})

(x - 2)^{\frac{1}{3}}

What is the tangent of f(x)=x^3-11x,\at x=3 ?
The tangent of f(x)=x^3-11x,\at x=3 is y=16x-54