SOLUTION: If f(x) = x^5 + Ax^2 - 2Ax + 2 and f(-1) = 4, find f(1).

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.

RELATED QUESTIONS
If f(x) = x^2 + Ax + B, f(3) = 20, and f(-1) = 4, find... (answered by josgarithmetic)

Given f(x) = (Ax+5)/(6x-2) If f(1) = 4, find... (answered by user_dude2008)

The function f(x) = ax^r satisfies f(2) = 1 and f(32) = 4. Find... (answered by robertb)

If f(x)=ax^2+bx+c Show that (f(x-h)-f(x))/h =... (answered by jsmallt9)

Please help me with this problem: Suppose f(x) = {{{ax^2}}} + bx +c. Find a, b, and c if (answered by stanbon)

Find the values of a and b that will make the function continuous every where. Justify. (answered by KMST)

Find the function f(x)=ax^3+bx^2+cx+d for which f(0)=-2, f(1)=5, f(-1)=3,... (answered by MathLover1)

given f(x) = ax^2 + bx + 5. Find a and b such that f(x + 1) - f(x) = 8x +... (answered by jim_thompson5910)

If f(x) = 5x +1 and g(x) = ax - 2, find 'a' such that (f +g)(5) =... (answered by josgarithmetic)