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

The tangent of f(x)=(x-1)^3,\at x=2 is y=3x-5

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

tangent of

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

y = 3 x - 5

Solution steps

Find the tangent point:

(2, 1)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 3

Find the line with slope m=

3

and passing through

(2, 1) : y = 3 x - 5

y = 3 x - 5

\int \frac{1}{x ( 2 a - x )} d x

\int_{1}^{l n (3)} \frac{1}{x} d x

\int 25 - 8 x + x^{2} d x

\frac{d}{d x} (2 x^{3} - 4 x - 8 x + 6)

\frac{1}{s ^{2}} (1 - e^{- 5 s})

What is the tangent of f(x)=(x-1)^3,\at x=2 ?
The tangent of f(x)=(x-1)^3,\at x=2 is y=3x-5