What is the tangent of f(x)=(x^3-4x)^8,\at x=2 ?

The tangent of f(x)=(x^3-4x)^8,\at x=2 is y=0

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tangent of f(x)=(x^3-4x)^8,\at x=2

tangent of

f (x) = (x^{3} - 4 x)^{8}, at x = 2

y = 0

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

f (x) = (x^{3} - 4 x)^{8} : \frac{df ( x )}{d x} = 8 (x^{3} - 4 x)^{7} (3 x^{2} - 4)

Compute the slope of

f (x) = (x^{3} - 4 x)^{8}

x = 2 : m = 0

Find the line with slope m=

0

and passing through

(2, 0) : y = 0

y = 0

f (x) = 3^{2 x}

(3 y^{2} - x y) d x + x^{2} d y = 0

\int 2 x^{2} 1 - x^{3} d x

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

\int (2 - 3 e^{x}) d x

What is the tangent of f(x)=(x^3-4x)^8,\at x=2 ?
The tangent of f(x)=(x^3-4x)^8,\at x=2 is y=0