If f(x+1)=x^2-4x, find the largest value of x such that f(x)=0.?

Answer 1

Because you are given the definition of f(x+1), plug x-1 in place of x and set the equation equal to 0.

(x-1)^2 -4(x-1) = 0
x^2 - 2x +1 - 4x +4 = 0
x^2 - 6x + 5 = 0
(x-5)(x-1) = 0
x = 1 or 5

The greatest value such that f(x) = 0 is 5.

Hope that helped.

Answer 2

In order to change f(x+1) to f(x) you need to replace (x+1) with something that will change x=1 to x. So f(x-1 +1) = f(x) does the trick. (We replaced the x in f(x+1) with (x-1)

So f(x) = (x-1)^2 -4(x-1) = x^2 -2x +1 -4x +4 = x^2 -6x +5
=(x-5)(x-1)
Thus x= 1 or x= 5
% is the correct answer

Answer 3

A different way to solve it: let x=y+1. Then we have

f(x) = f(y+1) = y^2-4y = y(y-4) = 0.

So y=0 or y=4, resulting in x=1 or x=5, the latter being the answer.

Answer 4

f(x)=(x-1)^2-4(x-1)

(x-1)(x-1-4)=0
x=1 or x=5

So the largest value of x is 5.

Answer 5

(X-1)(X-1-4)=0
X=1 OR X=5