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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 12

Find the line with slope m=

12

and passing through

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

y = 12 x - 3

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

\int_{- 3}^{3} ∣ x ∣ - (x^{2} - 6) d x

\int_{5}^{6} \frac{x}{x - 5} d x

\int \frac{x ^{2}}{x + 5} d x

\int \frac{x - 3}{x ^{3} + 3 x ^{2} + 2 x} d x

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