+1207 If \(\displaystyle{f(x)=x^{(x+1)}(x+2)^{(x+3)}}\), then find the value of \(f(0)+f(-1)+f(-2)+f(-3)\).
+129899 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 * 1 = 1
f(-2) = (-2)^(-2 + 1) * (-2 +2)^(-2 + 3) = (-2)^(-1)* (0)^(1) = 0
f(-3) = (-3)^(-3 + 1) * (-3 + 2)^(-3 + 3) = (-3)^(-2) * (-1)^0 = (1/9)*(1) = 1/9
The sum of these = 1 1/9 = 10/9
