What is the tangent of f(x)=x^3(3-x)^4,\at x=2 ?

The tangent of f(x)=x^3(3-x)^4,\at x=2 is y=-20x+48

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tangent of f(x)=x^3(3-x)^4,\at x=2

tangent of

f (x) = x^{3} (3 - x)^{4}, at x = 2

y = - 20 x + 48

Solution steps

Find the tangent point:

(2, 8)

Find the slope of

f (x) = x^{3} (3 - x)^{4} : \frac{df ( x )}{d x} = 3 x^{2} (3 - x)^{4} - 4 x^{3} (3 - x)^{3}

Compute the slope of

f (x) = x^{3} (3 - x)^{4}

x = 2 : m = - 20

Find the line with slope m=

- 20

and passing through

(2, 8) : y = - 20 x + 48

y = - 20 x + 48

x \to 100 lim (20 + 0.188 x)

\int y 6 + 12 y - 36 y^{2} d y

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

\int_{1}^{5} 6 x ln (x) d x

\int cos (x + n x) d x

What is the tangent of f(x)=x^3(3-x)^4,\at x=2 ?
The tangent of f(x)=x^3(3-x)^4,\at x=2 is y=-20x+48