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

The tangent of f(x)=x(4-x)^3,\at x=2 is y=-16x+48

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

tangent of

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

y = - 16 x + 48

Solution steps

Find the tangent point:

(2, 16)

Find the slope of

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

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 = - 16

Find the line with slope m=

- 16

and passing through

(2, 16) : y = - 16 x + 48

y = - 16 x + 48

What is the tangent of f(x)=x(4-x)^3,\at x=2 ?
The tangent of f(x)=x(4-x)^3,\at x=2 is y=-16x+48