What is the tangent of f(x)=(9x^9-9)(8-8x^8),\at x=0 ?

The tangent of f(x)=(9x^9-9)(8-8x^8),\at x=0 is y=-72

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tangent of f(x)=(9x^9-9)(8-8x^8),\at x=0

tangent of

f (x) = (9 x^{9} - 9) (8 - 8 x^{8}), at x = 0

y = - 72

Solution steps

Find the tangent point:

(0, - 72)

Find the slope of

f (x) = (9 x^{9} - 9) (8 - 8 x^{8}) : \frac{df ( x )}{d x} = 648 x^{8} - 1224 x^{16} + 576 x^{7}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 0

Find the line with slope m=

0

and passing through

(0, - 72) : y = - 72

y = - 72

\int - \frac{1}{sin ( x )} d x

\int_{0}^{4} x e^{- a x} d x

e^{8 x}

\int x^{2} 81 - x^{2} d x

f (x) = x 9 - x

What is the tangent of f(x)=(9x^9-9)(8-8x^8),\at x=0 ?
The tangent of f(x)=(9x^9-9)(8-8x^8),\at x=0 is y=-72