Question 919925: If f(x) = x^5 + Ax^2 - 2Ax + 2 and f(-1) = 4, find f(1).
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your equation is:
f(x) = x^5 + ax^2 - 2ax + 2
you also know that f(-1) = 4
replace x with -1 in your original equation and you get:
f(-1) = (-1)^5 + a(-1)^2 - 2a(-1) + 2 which becomes:
f(-1) = -1 + a + 2a + 2 which becomes:
f(-1) = 3a + 1
since f(-1) = 4, this means that:
3a + 1 = 4
subtract 1 from both sides of that equation to get:
3a = 3 which results in:
a = 1
that's your solution.
to confirm, replace a in the original equation with 1.
the original equation becomes:
x^5 + x^2 - 2x + 2
f(-1) = (-1)^5 + (-1)^2 - 2(-1) + 2 which becomes:
f(-1) = -1 + 1 + 2 + 2 which becomes:
f(-1) = 4
this agrees with what you were given that f(-1) = 4, so the solution of a = 1 is good.
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