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

The tangent of f(x)=3x^3,(-1,-3) is y=9x+6

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

tangent of

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

y = 9 x + 6

Solution steps

Find the slope of

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

Compute the slope of

f (x) = 3 x^{3}

(- 1, - 3) : m = 9

Find the line with slope m=

9

and passing through

(- 1, - 3) : y = 9 x + 6

y = 9 x + 6

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

\frac{d}{d x} (e^{π})

y = - 8 - (\frac{x}{7})^{\frac{5}{2}}

\frac{d}{d x} (4^{3 s i n (x) - e^{x}})

(3 P + 8)^{7}

What is the tangent of f(x)=3x^3,(-1,-3) ?
The tangent of f(x)=3x^3,(-1,-3) is y=9x+6