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}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 3

Find the line with slope m=

3

and passing through

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

y = 3 x - 5

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