SOLUTION: Let(f)=ax^5+bx^3+cx+9 were a,b,and c are constants. If f(6)=17, determine the value of f(-6)

Algebra ->  Functions -> SOLUTION: Let(f)=ax^5+bx^3+cx+9 were a,b,and c are constants. If f(6)=17, determine the value of f(-6)      Log On


   

Question 407705: Let(f)=ax^5+bx^3+cx+9 were a,b,and c are constants.
If f(6)=17, determine the value of f(-6)
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Note that all the x exponents (except for the 9) are odd, which means that x%5E5+=+-%28-x%29%5E5. Since f(6) = 17, then f(6) - 9 = 8, and

a%2A6%5E5+%2B+b%2A6%5E3+%2B+c%2A6+=+8

Since the function is odd, then

a%2A%28-6%29%5E5+%2B+b%2A%28-6%29%5E3+%2B+c%2A%28-6%29+=+-8

Adding 9 to obtain the value of f(-6) we get f(-6) = 1.