1995 AHSME Problems/Problem 14

Revision as of 16:37, 18 August 2011 by Talkinaway (talk | contribs) (Solution)

Problem

If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$, then $f(3) =$

$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$

Solution

$f(-x) = a(-x)^4 - b(-x)^2 - x + 5$f(-x) = ax^4 - bx^2 - x + 5

$f(-x) = (ax^4 + bx^2 + x + 5) - 2x$

$f(-x) =  f(x) - 2x$.

Thus $f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}$.

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions