Given Information:
The provided function is,
f(x,y,z)=xy+xz+x2y
Formula used:
A real-valued function, f, of x,y,z,… is a rule for manufacturing a new number, written f(x,y,z,...) from the values of a sequence of independent variables (x,y,z,…).
Calculation:
Consider the provided function,
f(x,y,z)=xy+xz+x2y
Now, to evaluate f(0,1,1), substitute 0 for x, 1 for y and 1 for z in the above function to get,
f(0,1,1)=01+0(1)+02(1)=0
Thus, f(0,1,1)=0
To evaluate f(2,1,1), substitute 2 for x, 1 for y and 1 for z in the above function to get,
f(2,1,1)=21+2(1)+22(1)=23+4=143
Thus, f(2,1,1)=143.
To evaluate f(−1,1,−1), substitute −1 for x, 1 for y and −1 for z in the above function to get,
f(−1,1,−1)=−11+(−1)(−1)+(−1)2(1)=−12+1=12
Thus, f(−1,1,−1)=12.
To evaluate f(z,z,z), substitute z for x, z for y and z for z in the above function to get,
f(z,z,z)=zz+z(z)+z2(z)=zz+z2+z3=11+z+z3
Thus, f(z,z,z)=11+z+z3.
To evaluate f(x+h,y+k,z+l), substitute x+h for x, y+k for y and z+l for z in the above function to get,
f(z+h,y+,z+l)=x+hy+k+(x+h)(z+l)+(x+h)2(y+k)
Thus, f(x+h,y+k,z+l)=x+hy+k+(x+h)(z+l)+(x+h)2(y+k).
Thus, for the provided function f(0,1,1)=0, f(2,1,1)=143, f(−1,1,−1)=12, f(z,z,z)=11+z+z3 and f(x+h,y+k,z+l)=x+hy+k+(x+h)(z+l)+(x+h)2(y+k).