What is the tangent of f(x)=x^2(1-x)^2,\at x=2 ?

The tangent of f(x)=x^2(1-x)^2,\at x=2 is y=12x-20

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tangent of f(x)=x^2(1-x)^2,\at x=2

tangent of

f (x) = x^{2} (1 - x)^{2}, at x = 2

y = 12 x - 20

Solution steps

Find the tangent point:

(2, 4)

Find the slope of

f (x) = x^{2} (1 - x)^{2} : \frac{df ( x )}{d x} = 4 x^{3} - 6 x^{2} + 2 x

Compute the slope of

f (x) = x^{2} (1 - x)^{2}

x = 2 : m = 12

Find the line with slope m=

12

and passing through

(2, 4) : y = 12 x - 20

y = 12 x - 20

x \to 0 lim (\frac{( x - 1 ) ^{2} + 1}{x})

\frac{( s + 1 ) ^{2}}{( s - 1 ) ^{3}}

\frac{\partial}{\partial x} (2 x^{2} y - 3)

\int x^{2} lo g_{10} (x) d x

f (x) = \frac{x ^{2} - 3 x + 2}{x ^{2} - 1}

What is the tangent of f(x)=x^2(1-x)^2,\at x=2 ?
The tangent of f(x)=x^2(1-x)^2,\at x=2 is y=12x-20