What is the tangent of f(x)=(x^3-9x)^{12},\at x=3 ?

The tangent of f(x)=(x^3-9x)^{12},\at x=3 is y=0

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tangent of f(x)=(x^3-9x)^{12},\at x=3

tangent of

f (x) = (x^{3} - 9 x)^{12}, at x = 3

y = 0

Solution steps

Find the tangent point:

(3, 0)

Find the slope of

f (x) = (x^{3} - 9 x)^{12} : \frac{df ( x )}{d x} = 12 (x^{3} - 9 x)^{11} (3 x^{2} - 9)

Compute the slope of

f (x) = (x^{3} - 9 x)^{12}

x = 3 : m = 0

Find the line with slope m=

0

and passing through

(3, 0) : y = 0

y = 0

\frac{9}{t ^{4}}

\int_{- 5}^{5} (4 - ∣ x ∣) d x

y = 8 csc (2 x^{4} - 7 x + 3)

\frac{d}{d x} (5 x^{5} f (x))

What is the tangent of f(x)=(x^3-9x)^{12},\at x=3 ?
The tangent of f(x)=(x^3-9x)^{12},\at x=3 is y=0