1) Given that f(x,y)=xy/x^2+y^2, hence show that f(xy) = f(yx)
2) use the mean value theorem to determine the value of c, given that f(x) = cos5x
To show that f(xy) = f(yx), we need to substitute these values into the expression for f(x, y) and show that they are equal.
The expression for f(x, y) is given as f(x, y) = xy / (x^2 + y^2).
1) Substitute xy into the expression:
f(xy) = (xy) / ((xy)^2 + y^2)
2) Simplify the denominator:
(xy)^2 + y^2 = x^2y^2 + y^2 = y^2(x^2 + 1)
3) Substitute yx into the expression:
f(yx) = (yx) / ((yx)^2 + x^2)
4) Simplify the denominator:
(yx)^2 + x^2 = y^2x^2 + x^2 = x^2(y^2 + 1)
Now, we compare f(xy) and f(yx):
f(xy) = (xy) / (y^2(x^2 + 1))
f(yx) = (yx) / (x^2(y^2 + 1))
We can see that these expressions are equal because the numerator and denominator of both expressions are the same, just rearranged. Therefore, we have shown that f(xy) = f(yx).
-------------------------------------------------------------------------------------------------------------------------
To determine the value of c using the mean value theorem for the function f(x) = cos(5x), we need to consider the interval [a, b] where the theorem applies. The mean value theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists a value c in (a, b) such that the derivative of the function evaluated at c is equal to the average rate of change of the function over the interval [a, b].
For our function f(x) = cos(5x), we can choose any interval [a, b] on which the function is continuous and differentiable. Let's consider the interval [0, π/5] as an example. In this interval, the function is both continuous and differentiable.
1) Compute the derivative of f(x):
f'(x) = -5sin(5x)
2) Compute the average rate of change of f(x) over the interval [0, π/5]:
Average rate of change = (f(b) - f(a)) / (b - a) = (f(π/5) - f(0)) / (π/5 - 0) = (cos(π) - cos(0)) / (π/5) = (1 - 1) / (π/5) = 0
3) Set the derivative equal to the average rate of change:
-5sin(5c) = 0
To determine the value of c, we solve for sin(5c) = 0:
sin(5c) = 0
The solutions to this equation are c = 0, π/5, 2π/5, 3π/5, 4π/5, etc.
So, using the mean value theorem, we have found that for the function f(x) = cos(5x) and the interval [0, π/5], the value of c can be any of the solutions c = 0, π/5, 2π/5, 3π/5, 4π/5, etc.