What is the tangent of f(x)=(x^3-16x)^8,\at x=4 ?

The tangent of f(x)=(x^3-16x)^8,\at x=4 is y=0

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tangent of f(x)=(x^3-16x)^8,\at x=4

tangent of

f (x) = (x^{3} - 16 x)^{8}, at x = 4

y = 0

Solution steps

Find the tangent point:

(4, 0)

Find the slope of

f (x) = (x^{3} - 16 x)^{8} : \frac{df ( x )}{d x} = 8 (x^{3} - 16 x)^{7} (3 x^{2} - 16)

Compute the slope of

f (x) = (x^{3} - 16 x)^{8}

x = 4 : m = 0

Find the line with slope m=

0

and passing through

(4, 0) : y = 0

y = 0

\int_{0}^{\infty} e^{- b x} d x

n = 0 \sum \infty (\frac{3}{4})^{n}

x^{4} + x

\int e^{4 - x^{2}} d x

\int \frac{sin ( x )}{cos ^{3} ( x ) + cos ( x )} d x

What is the tangent of f(x)=(x^3-16x)^8,\at x=4 ?
The tangent of f(x)=(x^3-16x)^8,\at x=4 is y=0