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
1) To show that f(xy) = f(yx) for the function f(x,y) = xy/(x^2 + y^2), we need to substitute xy in place of x and yx in place of y and verify that the equation holds true.
Let's start by computing f(xy):
f(xy) = (xy)/(x^2 + y^2)
Now, let's compute f(yx):
f(yx) = (yx)/(y^2 + x^2)
Notice that these expressions look almost the same, except that the terms in the denominator are switched. To show that f(xy) = f(yx), we need to prove that these expressions are equal.
To do this, we need to simplify the expressions and compare them:
f(xy) = (xy)/(x^2 + y^2)
= xy/(yx + x^2)
= yx/(x^2 + yx) [rearranging terms]
We can observe that this expression matches f(yx). Therefore, we have:
f(xy) = f(yx)
Thus, we have shown that f(xy) = f(yx) for the given function f(x,y) = xy/(x^2 + y^2).
2) To find the value of c using the Mean Value Theorem for the function f(x) = cos(5x), we need to first state the theorem and then apply it to solve for c.
The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the open interval (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
In this case, we have f(x) = cos(5x), so we need to find a and b to apply the theorem. Assuming we want to find the value of c in the interval [a, b], we need to choose appropriate values for a and b.
Let's assume a = 0 and b = π/10 (or any other suitable values depending on the requirements). Then, we can compute f(a) and f(b):
f(a) = f(0) = cos(5(0)) = cos(0) = 1
f(b) = f(π/10) = cos(5(π/10)) = cos(π/2) = 0
Now, we need to find f'(c) using the Mean Value Theorem:
f'(c) = (f(b) - f(a))/(b - a)
= (0 - 1)/(π/10 - 0)
= -1/(π/10)
= -10/π
Thus, we have found that f'(c) = -10/π using the Mean Value Theorem. This gives us the value of the derivative at c.