What is the tangent of f(x)=x^3-9x,(3,0) ?

The tangent of f(x)=x^3-9x,(3,0) is y=18x-54

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tangent of f(x)=x^3-9x,(3,0)

tangent of

f (x) = x^{3} - 9 x, (3, 0)

y = 18 x - 54

Solution steps

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

(3, 0) : m = 18

Find the line with slope m=

18

and passing through

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

y = 18 x - 54

t \frac{d y}{d t} = t^{3} + 4 t^{3} y

\int (x^{2} + 3) 5 x d x

f (x) = (3 x^{3} + 2) (x^{4} - 2 x)

\frac{d y}{d x} x + y = (x y)^{\frac{3}{2}}

\int \frac{1}{1 + 8 x} d x

What is the tangent of f(x)=x^3-9x,(3,0) ?
The tangent of f(x)=x^3-9x,(3,0) is y=18x-54