What is the tangent of f(x)=(x^3-16x)^{10},(4,0) ?

The tangent of f(x)=(x^3-16x)^{10},(4,0) is y=0

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tangent of f(x)=(x^3-16x)^{10},(4,0)

tangent of

f (x) = (x^{3} - 16 x)^{10}, (4, 0)

y = 0

Solution steps

Find the slope of

f (x) = (x^{3} - 16 x)^{10} : \frac{df ( x )}{d x} = 10 (x^{3} - 16 x)^{9} (3 x^{2} - 16)

Compute the slope of

f (x) = (x^{3} - 16 x)^{10}

(4, 0) : m = 0

Find the line with slope m=

0

and passing through

(4, 0) : y = 0

y = 0

\int_{0}^{\frac{π}{3}} cos^{2} (x) d x

\int \frac{ln ( u )}{u} d u

f^{9}

n = 1 \sum \infty \frac{5}{n ^{2}}

\int (2 x + 1)^{3} 2 d x

What is the tangent of f(x)=(x^3-16x)^{10},(4,0) ?
The tangent of f(x)=(x^3-16x)^{10},(4,0) is y=0