What is the tangent of f(x)=(x^3-25x)^8,(-5,0) ?

The tangent of f(x)=(x^3-25x)^8,(-5,0) is y=0

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tangent of f(x)=(x^3-25x)^8,(-5,0)

tangent of

f (x) = (x^{3} - 25 x)^{8}, (- 5, 0)

y = 0

Solution steps

Find the slope of

f (x) = (x^{3} - 25 x)^{8} : \frac{df ( x )}{d x} = 8 (x^{3} - 25 x)^{7} (3 x^{2} - 25)

Compute the slope of

f (x) = (x^{3} - 25 x)^{8}

(- 5, 0) : m = 0

Find the line with slope m=

0

and passing through

(- 5, 0) : y = 0

y = 0

\int \frac{2 x}{1 - x} d x

\int \frac{- 3 x}{( x - 1 ) ^{2} + 4} d x

y^{^{'''}} - y^{^{''}} + y^{^{'}} + 3 y = 0

\frac{d y}{d t} = 3 y + y^{4}, y (0) = 1

\frac{d}{d x} (- 20 x^{3} + 36 x^{2} - 12 x - 4)

What is the tangent of f(x)=(x^3-25x)^8,(-5,0) ?
The tangent of f(x)=(x^3-25x)^8,(-5,0) is y=0