Compute the following values for the given function.
f(x, y) = 5xye^(x^2+y^2)
a. f(0,0)= ?
b. f(0,1)= ?
c. f(1,1)= ?
d. f(-1,-1)= ?
To compute the values for the given function, substitute the given values of x and y into the function and simplify accordingly.
a. f(0,0):
Start with the function f(x, y) = 5xye^(x^2+y^2). Plug in x=0 and y=0:
f(0,0) = 5(0)(0)e^(0^2+0^2)
Since any number raised to the power of 0 is equal to 1, we have:
f(0,0) = 5(0)(0)(1)
Any number multiplied by 0 is equal to 0, so:
f(0,0) = 0
Therefore, f(0,0) = 0.
b. f(0,1):
Using the same function f(x, y) = 5xye^(x^2+y^2), substitute x=0 and y=1:
f(0,1) = 5(0)(1)e^(0^2+1^2)
Again, since 0 raised to any power is equal to 0, we have:
f(0,1) = 5(0)(1)(e^1)
Simplifying further:
f(0,1) = 0
Therefore, f(0,1) = 0.
c. f(1,1):
For f(1,1), substitute x=1 and y=1 into the function:
f(1,1) = 5(1)(1)e^(1^2+1^2)
Simplifying the exponents:
f(1,1) = 5(1)(1)(e^2)
Therefore, f(1,1) = 5e^2.
d. f(-1,-1):
To compute f(-1,-1), substitute x=-1 and y=-1 into the function:
f(-1,-1) = 5(-1)(-1)e^((-1)^2+(-1)^2)
Simplifying the exponents:
f(-1,-1) = 5(-1)(-1)(e^2)
Therefore, f(-1,-1) = 5e^2.