Question

What is the equation of the line tangent to `f(x)=(x-1)^3` at `x=2`?

Answer 1

`y = 3x-5`

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent so the product of their gradients is `-1`

We have:

`f(x)=(x-1)^3`

Then differentiating wrt `x`, and applying the chain rule gives us:

`f'(x) = 3(x-1)^2(1)`

So when `x=2`, then;

`f(2) \ \= (2-1)^3 \ \ = 1`
`f'(2) = 3(2-1)^2 = 3`

So the tangent passes through `(2,1)` and has gradient `m_T=3`, so using the point/slope form `y-y_1=m(x-x_1)` the equation we seek are;

`y - 1 = 3(x-2)`
`:. y - 1 = 3x-6`
`:. y = 3x-5`

We can confirm this solution is correct graphically: