What is the tangent of ((x^3-16x)^{12}),\at x=-4 ?

The tangent of ((x^3-16x)^{12}),\at x=-4 is y=0

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tangent of ((x^3-16x)^{12}),\at x=-4

tangent of

((x^{3} - 16 x)^{12}), at x = - 4

y = 0

Solution steps

Find the tangent point:

(- 4, 0)

Find the slope of

y = (x^{3} - 16 x)^{12} : \frac{d y}{d x} = 12 (x^{3} - 16 x)^{11} (3 x^{2} - 16)

Compute the slope of

y = (x^{3} - 16 x)^{12}

x = - 4 : m = 0

Find the line with slope m=

0

and passing through

(- 4, 0) : y = 0

y = 0

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What is the tangent of ((x^3-16x)^{12}),\at x=-4 ?
The tangent of ((x^3-16x)^{12}),\at x=-4 is y=0