If \(\displaystyle{f(x)=x^{(x+1)}(x+2)^{(x+3)}}\), then find the value of \(f(0)+f(-1)+f(-2)+f(-3)\).
+128474 f(0) = 0^(0 + 1)*(0 +2)^(0 + 3) = 0^1 * 2^3 = 0
f(-1) = (-1)^(-1+1)*(-1+ 2)^(-1+3) = (-1)^0 * (1)^2 = 1
f(-2) = (-2)^(-2 + 1) * ( -2 + 2)^(-2 + 3) = (-2)^(-1) * (0)^(1) = 0
f(-3) = (-3)^(-3 + 1) * ( -3 + 2)^(0) = (-3)^(-2) * (-1)^0 = 1/(-3)^2 = 1/9
So
0 + 1 + 0 +1/9 =
1 + 1/9 =
10 / 9
