1982 AHSME Problems/Problem 12


Let $f(x) = ax^7+bx^3+cx-5$, where $a,b$ and $c$ are constants. If $f(-7) = 7$, then $f(7)$ equals

$\text {(A)} -17 \qquad  \text {(B)} -7 \qquad \text {(C)} 14 \qquad  \text {(D)} 21\qquad  \text {(E)} \text{not uniquely determined}$


$f(x)$ is an odd function shifted down 5 units. Thus, it can be written as $f(x)=g(x)-5$ where $g(x)=ax^7+bx^3+cx$. Thus: $f(-7)=g(-7)-5=7$ and $g(-7)=12$. Using this and the fact $g(x)$ is odd, we can evaluate $f(7)$, which is:

\[f(7) = g(7)-5 = -g(-7)-5 = -12-5 = -17\]

Therefore, the answer is $\boxed{ \textbf{A}}$.