What is the tangent of f(x)=x^2(4-x)^2,\at x=1 ?

The tangent of f(x)=x^2(4-x)^2,\at x=1 is y=12x-3

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tangent of f(x)=x^2(4-x)^2,\at x=1

tangent of

f (x) = x^{2} (4 - x)^{2}, at x = 1

y = 12 x - 3

Solution steps

Find the tangent point:

(1, 9)

Find the slope of

f (x) = x^{2} (4 - x)^{2} : \frac{df ( x )}{d x} = 4 x^{3} - 24 x^{2} + 32 x

Compute the slope of

f (x) = x^{2} (4 - x)^{2}

x = 1 : m = 12

Find the line with slope m=

12

and passing through

(1, 9) : y = 12 x - 3

y = 12 x - 3

\int_{0}^{1} \frac{1}{1 - x ^{2}} d x

x \to 0 lim (\frac{3}{e ^{x} + 1})

\frac{\partial}{\partial x} (6 x^{2} + 7 y)

\frac{d}{d x} (∣ x ∣^{3} - x^{2})

x \to \infty lim (\frac{12 x ^{3} - 4 x + 1}{1 + 3 x + 2 x ^{3}})

What is the tangent of f(x)=x^2(4-x)^2,\at x=1 ?
The tangent of f(x)=x^2(4-x)^2,\at x=1 is y=12x-3