2580.41 – A Seventh Power Equation


Let f(x)=ax7+bx3+cx5f(x) = ax^7 + bx^3 + cx - 5, where a, b,a, b, and cc are constants.

If f(7)=7f(-7) = 7, then what is f(7)f(7)?

  1. 17-17

  2. 7-7

  3. 1414

  4. 2121

  5. not uniquely determined


Solution

Since f(x) = ax7 + bx3 + cx –5,f(x) = ax^7 + bx^3 + cx – 5, and f(7)=7f(-7) = 7 then

f(7)= a(7)7 + b(7)3 + c(7)5,7= a(7)7 + b(7)3 + c(7)5,7=a77  b73  c(7)57=(a77b73c(7)5)107=f(7)10.\begin{aligned}f(-7) &= a(-7)^7 + b(-7)^3 + c(-7) – 5, \\7 &= a(-7)^7 + b(-7)^3 + c(-7) - 5, \\7 &= -a7^7 - b7^3 - c(7)- 5 \\7 &= -(a7^7 - b7^3 - c(7)- 5) - 10 \\7 &= -f(7) - 10.\end{aligned}

So f(7)=107=17.f(7) = -10 - 7 = -17. The answer is (a).