What is the tangent of f(x)=x^3-12x,\at x=1 ?

The tangent of f(x)=x^3-12x,\at x=1 is y=-9x-2

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tangent of f(x)=x^3-12x,\at x=1

tangent of

f (x) = x^{3} - 12 x, at x = 1

y = - 9 x - 2

Solution steps

Find the tangent point:

(1, - 11)

Find the slope of

f (x) = x^{3} - 12 x : \frac{df ( x )}{d x} = 3 x^{2} - 12

Compute the slope of

f (x) = x^{3} - 12 x

x = 1 : m = - 9

Find the line with slope m=

- 9

and passing through

(1, - 11) : y = - 9 x - 2

y = - 9 x - 2

\frac{d}{d x} (\frac{x - 13}{x - 13})

(1.2) (- 2.14)

\frac{d}{d x} (\frac{sinh ( x )}{cosh ( x )})

\frac{d}{d x} (\frac{x - 10}{x - 10})

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

What is the tangent of f(x)=x^3-12x,\at x=1 ?
The tangent of f(x)=x^3-12x,\at x=1 is y=-9x-2