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

The tangent of f(x)=x(3-x)^2,\at x=2 is y=-3x+8

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

tangent of

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

y = - 3 x + 8

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

f (x) = x (3 - x)^{2} : \frac{df ( x )}{d x} = 3 x^{2} - 12 x + 9

Compute the slope of

f (x) = x (3 - x)^{2}

x = 2 : m = - 3

Find the line with slope m=

- 3

and passing through

(2, 2) : y = - 3 x + 8

y = - 3 x + 8

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

x \to 2 lim (a x^{2} - 2 b x + 1)

\int \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}} d x

\frac{\partial}{\partial x} (sin^{2} (k x))

y^{^{'}} - \frac{9}{x} y = \frac{( y ^{5} )}{x ^{8}}

What is the tangent of f(x)=x(3-x)^2,\at x=2 ?
The tangent of f(x)=x(3-x)^2,\at x=2 is y=-3x+8