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}

Compute the slope of

f (x) = x (4 - x)^{3}

x = 2 : m = - 16

Find the line with slope m=

- 16

and passing through

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

y = - 16 x + 48

f (t) = 3 sin (4 t) - 2 e^{2 t}

h (t) = 6 ln (t^{8})

\frac{d}{d x} (144 - x)

\int (5 - x^{3})^{4} x^{2} d x

\frac{d}{d x} (x^{2} 2 x + 3)

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