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

The tangent of f(x)=x^4(3-x)^2,\at x=1 is y=12x-8

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

tangent of

f (x) = x^{4} (3 - x)^{2}, at x = 1

y = 12 x - 8

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = x^{4} (3 - x)^{2} : \frac{df ( x )}{d x} = 6 x^{5} - 30 x^{4} + 36 x^{3}

Compute the slope of

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

x = 1 : m = 12

Find the line with slope m=

12

and passing through

(1, 4) : y = 12 x - 8

y = 12 x - 8

\frac{d y}{d x} = 200 - \frac{3 y}{100 - 10 x}

f (x) = 4 x^{2} + 7 x

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

\frac{d}{d x} ((4 x)^{l n (4 x)})

2 5^{\frac{1}{3}} x^{\frac{2}{3}}

What is the tangent of f(x)=x^4(3-x)^2,\at x=1 ?
The tangent of f(x)=x^4(3-x)^2,\at x=1 is y=12x-8