Question

Given that `f(x)=(x-2)^5+2(x-2)^4+3(x-2)^3+4(x-2)^2+5(x-2)+6`. If `f(x)=a(color(red)(x+2))^5+b(x-2)^4+c(x-2)^3+d(x-2)^2+e(x-2)+f`. Find the values of `a,b,c,d,e,f` ?

Note that the question is `a(x+2)^5` and not `a(x-2)^5` , which makes me have no clue how to solve it...

Answer 1

See below.

Explanation:

Calling

`f(x)=(x-2)^5+2(x-2)^4+3(x-2)^3+4(x-2)^2+5(x-2)+6`
`g(x)=a_0(x+2)^5+b_0(x-2)^4+c_0(x-2)^3+d_0(x-2)^2+e_0(x-2)+f_0`
we have

#{(f(2)=g(2)->6-a_0 4^5-f_0), (d/(dx)f(2)=d/(dx)g(2)->5 - 5*4^4 a_0 - e_0 = 0), (d^2/(dx^2)f(2)=d^2/(dx^2)g(2)->8 - 1280 a_0 - 2 d_0 = 0), (d^3/(dx^3)f(2)=d^3/(dx^3)g(2)->18 - 960 a_0 - 6 c_0 = 0), (d^4/(dx^4)f(2)=d^4/(dx^4)g(2)-> 48 - 480 a_0 - 24 b_0 = 0), (d^5/(dx^5)f(2)=d^5/(dx^5)g(2)->120 - 120 a_0 = 0):}#

Solving for `a_0,b_0,c_0,d_0,e_0,f_0` we obtain

`a_0=1,b_0=-18,c_0=-157,d_0=-636,e_0=-1275,f_0=-1018`