How do you find the equation of the tangent line to the curve f(x)=(x−1)3 at the point where x = 2?

Question

uvredio0of6 · Accepted Answer

To find the equation in the form y = mx + c , where m represents the gradient and c, the y-intercept.
The value of f'(2) will give m and f(2) will assist in finding c.
differentiate using the chain rule

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

hence

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

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

gradient of tangent
and

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

is tangent point
partial equation of tangent is y = 3x + c , and using (2,1 )
gives :

1 = 3 (2) + c \to c = - 5

thus equation of tangent is : y = 3x - 5

How do you find the equation of the

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