Question

How do you find the equation of a line tangent to a graph `f(x)=x(1-2x)^3` at the point (1,-1) is?

Answer 1

By the Linear Approximation Method, the tangent line to a curve is:

`f_T(x) = f(a) + f'(a)(x-a)`

You have `a = 1` since `a` is substituted in for `x` and you have `(1,-1)` as a coordinate of interest.

`f_T(x) = f(1) + f'(1)(x-1)`

`f(x) = overbrace(x)^(g(x))overbrace((1-2x)^3)^(h(x))`

`f'(x) = overbrace(x)^(g(x))overbrace((3(1-2x)^2*-2))^(h'(x)) + overbrace((1-2x)^3)^(h(x))*overbrace(1)^(g'(x))`

(Product Rule, Power Rule, Chain Rule)

`= -6x(1-2x)^2+(1-2x)^3`

`f_T(x) = (1)(1-2(1))^3`
`+ [-6(1)(1-2(1))^2+(1-2(1))^3]*(x-1)`

`f_T(x) = -1 + [-6-1]*(x-1)`

`f_T(x) = overbrace(-1)^(f(a)) overbrace(- 7)^(+ f'(a))overbrace((x-1))^((x-a))`

or

`f_T(x) = -1 - 7x + 7`

`= -7x+6`

You can see the result here.