Find the critical points of the following function. f(x) = 8x/(x^2+16)

To find the critical points of the function f(x) = 8x/(x^2+16), we first need to find its derivative.

The derivative of a quotient can be found using the quotient rule, which states that the derivative of u/v is (v * du/dx - u * dv/dx) / v^2.

Let's set u = 8x and v = x^2 + 16.

The derivative of u with respect to x, du/dx, is 8.

The derivative of v with respect to x, dv/dx, is 2x.

Substituting these into the quotient rule gives us:

f'(x) = ( (x^2 + 16) * 8 - 8x * 2x ) / (x^2 + 16)^2 = ( 8x^2 + 128 - 16x^2 ) / (x^4 + 32x^2 + 256) = ( -8x^2 + 128 ) / (x^4 + 32x^2 + 256)

The critical points of the function are the values of x that make the derivative equal to zero or undefined.

Setting the derivative equal to zero gives us:

-8x^2 + 128 = 0 x^2 = 16 x = ±4

The derivative is undefined when the denominator is zero. However, the equation x^4 + 32x^2 + 256 = 0 has no real solutions, so there are no critical points from the derivative being undefined.

Therefore, the critical points of the function are x = -4 and x = 4.