If f(x)=x^x+1(x+2)^x+3, then f(0)+f(-1) +f(-2)+f(-3)=

1996 AMC 12 Problem 6

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f(x)=x^{x+1}(x+2)^{x+3},

then

\begin{gathered} f(0)+f(-1)\\ {}+f(-2)+f(-3)= \end{gathered}

-\frac89

0

\frac89

1

\frac{10}9

Evaluate the four function values separately, paying attention to zero exponents

The terms at

x=0

and

x=-2

vanish before any problematic power is needed

Direct substitution gives

\begin{aligned} f(0)&=0,\\ f(-1)&=1,\\ f(-2)&=0,\\ f(-3)&=\frac19. \end{aligned}

Their sum is

\frac{10}{9},

so the correct answer is E.