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

The tangent of f(x)=x^4(3-x)^3,\at x=1 is y=20x-12

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

tangent of

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

y = 20 x - 12

Solution steps

Find the tangent point:

(1, 8)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 20

Find the line with slope m=

20

and passing through

(1, 8) : y = 20 x - 12

y = 20 x - 12

\frac{12 x}{x ^{2} - 4}

\frac{d}{d x} (4 cos (x) + 4 x sin (x))

\frac{d}{d x} (\frac{sin ( 12 x )}{e ^{7 x}})

\int_{0}^{\frac{π}{6}} 4 sec^{3} (x) d x

\frac{d ^{3}}{d x ^{3}} (\frac{1}{x - 1})

What is the tangent of f(x)=x^4(3-x)^3,\at x=1 ?
The tangent of f(x)=x^4(3-x)^3,\at x=1 is y=20x-12