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

The tangent of f(x)=(x-1)(x-3)^2,\at x=4 is y=7x-25

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

tangent of

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

y = 7 x - 25

Solution steps

Find the tangent point:

(4, 3)

Find the slope of

f (x) = (x - 1) (x - 3)^{2} : \frac{df ( x )}{d x} = 3 x^{2} - 14 x + 15

Compute the slope of

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

x = 4 : m = 7

Find the line with slope m=

7

and passing through

(4, 3) : y = 7 x - 25

y = 7 x - 25

(1 + 2 y) d x - (4 - x) d y = 0

f (x) = 13 x

\frac{d}{d x} (arcsin (7 x + 1))

\int (8 x^{3}) d x

\int 4 x cos (2 x) d x

What is the tangent of f(x)=(x-1)(x-3)^2,\at x=4 ?
The tangent of f(x)=(x-1)(x-3)^2,\at x=4 is y=7x-25