What is the equation of the tangent line of f(x)=(x−1)3 at x=2?

Question

Egerlandsq0z · Accepted Answer

To find the equation in the form y = mx + c , where m represents the gradient and c , the y-intercept.
f'(2) is the value of the tangent gradient and f(2) is the value of the y-coordinate , given x-coordinate.
differentiate using the chain rule

\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)

f^{'} (x) = 3 {(x - 1)}^{2} \cdot \frac{d}{d x} (x - 1) = 3 {(x - 1)}^{2}

and

f^{'} (2) = 3 {(2 - 1)}^{2} = 3 = m

gradient of tangent
hence 'partial' equation is y = 3x + c
now

f (2) = {(3 - 2)}^{2} = 1 \to (2, 1)

is point on tangent
using (2,1) in

y = 3 x + c : 6 + c = 1 \to c = - 5

and equation of tangent is therefore : y = 3x - 5

What is the equation of the tangent line

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