What is the tangent of f(x)=(x^3-9x)^8,(-3,0) ?

The tangent of f(x)=(x^3-9x)^8,(-3,0) is y=0

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tangent of f(x)=(x^3-9x)^8,(-3,0)

tangent of

f (x) = (x^{3} - 9 x)^{8}, (- 3, 0)

y = 0

Solution steps

Find the slope of

f (x) = (x^{3} - 9 x)^{8} : \frac{df ( x )}{d x} = 8 (x^{3} - 9 x)^{7} (3 x^{2} - 9)

Compute the slope of

f (x) = (x^{3} - 9 x)^{8}

(- 3, 0) : m = 0

Find the line with slope m=

0

and passing through

(- 3, 0) : y = 0

y = 0

\frac{d y}{d x} + 3 y = 2 x

x^{3}, x, 0, 1

x \to 4 lim (\frac{9}{9 - ( - 3 )})

x \to 0 lim (x^{2} \cdot ln (x))

\frac{d}{d x} ((x - π)^{2})

What is the tangent of f(x)=(x^3-9x)^8,(-3,0) ?
The tangent of f(x)=(x^3-9x)^8,(-3,0) is y=0