What is the tangent of f(x)=(x-4)(3-x)^3,\at x=5 ?

The tangent of f(x)=(x-4)(3-x)^3,\at x=5 is y=-20x+92

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tangent of f(x)=(x-4)(3-x)^3,\at x=5

tangent of

f (x) = (x - 4) (3 - x)^{3}, at x = 5

y = - 20 x + 92

Solution steps

Find the tangent point:

(5, - 8)

Find the slope of

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

Compute the slope of

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

x = 5 : m = - 20

Find the line with slope m=

- 20

and passing through

(5, - 8) : y = - 20 x + 92

y = - 20 x + 92

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

\int sin (- 3 x) d x

t \to π lim (- 9 sin (- \frac{t}{2}))

\int - 3 x + 4 x d x

x^{- \frac{3}{4}}

What is the tangent of f(x)=(x-4)(3-x)^3,\at x=5 ?
The tangent of f(x)=(x-4)(3-x)^3,\at x=5 is y=-20x+92